<?xml version='1.0' encoding='UTF-8'?>
<questestinterop xmlns="http://www.imsglobal.org/xsd/ims_qtiasiv1p2" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.imsglobal.org/xsd/ims_qtiasiv1p2 http://www.imsglobal.org/xsd/ims_qtiasiv1p2p1.xsd">
<objectbank ident="D4">
<qtimetadata>
<qtimetadatafield><fieldlabel>bank_title</fieldlabel><fieldentry>Differential Equations -- D4</fieldentry></qtimetadatafield>
</qtimetadata>
<item ident="D4-3628" title="D4 | Using Laplace transforms to solve IVPs | ver. 3628"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By%7D%20=%20-2%20%5C,%20%7By''%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" alt="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7074" title="D4 | Using Laplace transforms to solve IVPs | ver. 7074"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20=%206%20%5C,%20%7By%7D%20-%209%20%5C,%20%7By'%7D%20-%203%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y''} = 6 \, {y} - 9 \, {y'} - 3 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%202%7D" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}" alt="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}" title="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}" data-latex="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s^{2} - 3 \, s + 2} - \frac{2}{s^{2} - 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%201%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{e^{\left(-2 \, s\right)}}{s - 2} + \frac{2}{s - 1} - \frac{2}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%20e%5E%7B%5Cleft(t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20e%5E%7Bt%7D" alt="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}" title="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}" data-latex="{y} = e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{t}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1525" title="D4 | Using Laplace transforms to solve IVPs | ver. 1525"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" alt="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%206%20%5C,%20%7By'%7D%20-%206%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" alt="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-0666" title="D4 | Using Laplace transforms to solve IVPs | ver. 0666"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" alt="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%203%20%5C,%20%7By''%7D%20+%2027%20%5C,%20%7By%7D%20+%2027%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-6" alt="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" alt="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B9%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5049" title="D4 | Using Laplace transforms to solve IVPs | ver. 5049"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8" alt="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8" title="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12%20%5C,%20%7By%7D%20+%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-8" alt="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8" title="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="-12 \, {y} + 12 \, \delta\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%204%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}" title="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}" alt="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}" title="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%204%7D%20-%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 4} - \frac{8}{s^{2} - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B2%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}" title="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s + 2} - \frac{e^{\left(-s\right)}}{s - 2} + \frac{2}{s + 2} - \frac{2}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}" title="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(2 \, t\right)} + 2 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2484" title="D4 | Using Laplace transforms to solve IVPs | ver. 2484"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" alt="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20=%2018%20%5C,%20%7By%7D%20-%2015%20%5C,%20%7By'%7D%20-%206%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}" alt="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}" title="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}" data-latex="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D" alt="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}" title="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}" data-latex="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8248" title="D4 | Using Laplace transforms to solve IVPs | ver. 8248"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" alt="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4%20%5C,%20%7By'%7D%20+%2016%20%5C,%20%7By%7D%20=%202%20%5C,%20%7By''%7D%20-%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2012" alt="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20=%20-%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" alt="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" title="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20+%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%208%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" title="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8476" title="D4 | Using Laplace transforms to solve IVPs | ver. 8476"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6" alt="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6" title="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%20-6%20%5C,%20%7By'%7D%20-%2024%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-6" alt="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6" title="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20=%20-%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}" alt="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}" title="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20-%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}" title="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5523" title="D4 | Using Laplace transforms to solve IVPs | ver. 5523"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1" alt="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1" title="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%20-2%20%5C,%20%7By''%7D%20-%2024%20%5C,%20%7By%7D%20+%2014%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%201" alt="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1" title="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="-8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}" alt="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}" title="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20+%20%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}" title="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2645" title="D4 | Using Laplace transforms to solve IVPs | ver. 2645"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" alt="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20=%20-27%20%5C,%20%7By%7D%20-%2081%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-2%20,%20y'(0)=%200" alt="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8829" title="D4 | Using Laplace transforms to solve IVPs | ver. 8829"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16" alt="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16" title="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16" data-latex="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?24%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20+%209%20%5C,%20%7By%7D%20+%206%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-16" alt="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16" title="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16" data-latex="24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" alt="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" alt="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20-%20%5Cfrac%7B16%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7629" title="D4 | Using Laplace transforms to solve IVPs | ver. 7629"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0" alt="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0" title="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0" data-latex="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20-%2012%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%205%20,%20y'(0)=%200" alt="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0" title="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0" data-latex="-12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)" alt="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)" title="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)" data-latex="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(2%20%5C,%20t%20-%204%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%205%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)" title="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)" data-latex="{y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1900" title="D4 | Using Laplace transforms to solve IVPs | ver. 1900"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" alt="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" title="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-27%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20+%2081%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" title="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-27 \, {y} - 3 \, {y''} + 81 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)" alt="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 4 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1931" title="D4 | Using Laplace transforms to solve IVPs | ver. 1931"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8" alt="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8" title="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8" data-latex="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8%20%5C,%20%7By%7D%20-%2032%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%20-2%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%208" alt="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8" title="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8" data-latex="-8 \, {y} - 32 \, \delta\left(t - 1\right) = -2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%204%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}" alt="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}" title="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" alt="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" title="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B16%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%204%7D%20+%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20-%204%7D" alt="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}" title="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= \frac{16 \, e^{\left(-s\right)}}{s^{2} - 4} + \frac{8}{s^{2} - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 2} + \frac{2}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" title="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3893" title="D4 | Using Laplace transforms to solve IVPs | ver. 3893"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6" alt="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27%20%5C,%20%7By%7D%20=%20-3%20%5C,%20%7By''%7D%20-%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%206" alt="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="27 \, {y} = -3 \, {y''} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B18%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{6}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + 2 \, \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5438" title="D4 | Using Laplace transforms to solve IVPs | ver. 5438"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9%20%5C,%20%7By'%7D%20+%206%20%5C,%20%7By%7D%20=%20-3%20%5C,%20%7By''%7D%20+%209%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-9 \, {y'} + 6 \, {y} = -3 \, {y''} + 9 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%202%7D" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}" alt="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}" title="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}" data-latex="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s^{2} - 3 \, s + 2} + \frac{3}{s^{2} - 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s - 1} + \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3}{s - 1} + \frac{3}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(t%20-%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7Bt%7D" alt="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}" title="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}" data-latex="{y} = 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(t - 1\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t\right)} - 3 \, e^{t}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-0079" title="D4 | Using Laplace transforms to solve IVPs | ver. 0079"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0" alt="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0" title="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27%20%5C,%20%7By%7D%20+%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%20-1%20,%20y'(0)=%200" alt="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0" title="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="27 \, {y} + 54 \, \mathrm{u}\left(t - 1\right) = -3 \, {y''} \hspace{2em} y(0)= -1 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B18%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \cos\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7828" title="D4 | Using Laplace transforms to solve IVPs | ver. 7828"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" alt="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12%20%5C,%20%7By%7D%20=%20-3%20%5C,%20%7By''%7D%20+%2036%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-5%20,%20y'(0)=%200" alt="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="12 \, {y} = -3 \, {y''} + 36 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{12 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%205%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2090" title="D4 | Using Laplace transforms to solve IVPs | ver. 2090"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1" alt="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1" title="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%2010%20%5C,%20%7By'%7D%20+%2012%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%206%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%201" alt="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1" title="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="0 = 10 \, {y'} + 12 \, {y} + 2 \, {y''} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 1">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}" alt="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}" title="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}" data-latex="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20+%20%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{1}{s^{2} + 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{1}{s + 3} + \frac{1}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20-%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}" title="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}" data-latex="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-2 \, t\right)} - e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3566" title="D4 | Using Laplace transforms to solve IVPs | ver. 3566"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-54%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20=%202%20%5C,%20%7By''%7D%20+%2018%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-54 \, \mathrm{u}\left(t - 3\right) = 2 \, {y''} + 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)" alt="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%209%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6420" title="D4 | Using Laplace transforms to solve IVPs | ver. 6420"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1" alt="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1" title="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-6%20%5C,%20%7By%7D%20+%209%20%5C,%20%7By'%7D%20-%206%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%201" alt="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1" title="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1" data-latex="-6 \, {y} + 9 \, {y'} - 6 \, \delta\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 1">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%202%7D" alt="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" title="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\frac{1}{s^{2} - 3 \, s + 2} = -\frac{1}{s - 1} + \frac{1}{s - 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}" alt="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}" title="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}" data-latex="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s + 2} + \frac{1}{s^{2} - 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{1}{s - 1} + \frac{1}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%20e%5E%7Bt%7D" alt="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}" title="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}" data-latex="{y} = -2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) + e^{\left(2 \, t\right)} - e^{t}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7278" title="D4 | Using Laplace transforms to solve IVPs | ver. 7278"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6" alt="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%20-2%20%5C,%20%7By''%7D%20+%204%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By'%7D%20+%206%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%206" alt="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="0 = -2 \, {y''} + 4 \, {y} + 2 \, {y'} + 6 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" alt="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" title="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20+%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" title="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8332" title="D4 | Using Laplace transforms to solve IVPs | ver. 8332"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9" alt="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9" title="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9" data-latex="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-24%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%204%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By''%7D%20+%202%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-9" alt="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9" title="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9" data-latex="-24 \, \delta\left(t - 3\right) = 4 \, {y} - 2 \, {y''} + 2 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -9">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" alt="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" title="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" data-latex="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20-%20%5Cfrac%7B9%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" title="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" data-latex="{y} = 4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8184" title="D4 | Using Laplace transforms to solve IVPs | ver. 8184"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0" alt="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0" title="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0" data-latex="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20-%203%20%5C,%20%7By%7D%20+%206%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%201%20,%20y'(0)=%200" alt="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0" title="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0" data-latex="-3 \, {y''} - 3 \, {y} + 6 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 1 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} + \frac{s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20%5Ccos%5Cleft(t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6709" title="D4 | Using Laplace transforms to solve IVPs | ver. 6709"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7" alt="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7" title="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7" data-latex="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?24%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By'%7D%20=%202%20%5C,%20%7By''%7D%20+%2014%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-7" alt="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7" title="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7" data-latex="24 \, {y} - 2 \, {y'} = 2 \, {y''} + 14 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -7">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%20s%20-%2012%7D%20=%20-%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} + s - 12} = -\frac{1}{7 \, {\left(s + 4\right)}} + \frac{1}{7 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}" alt="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}" title="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}" data-latex="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}" alt="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}" title="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}" data-latex="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B7%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%20s%20-%2012%7D%20-%20%5Cfrac%7B7%7D%7Bs%5E%7B2%7D%20+%20s%20-%2012%7D" alt="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}" title="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}" data-latex="\mathcal{L}\{y\}= -\frac{7 \, e^{\left(-3 \, s\right)}}{s^{2} + s - 12} - \frac{7}{s^{2} + s - 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 4} - \frac{e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 4} - \frac{1}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}" title="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}" data-latex="{y} = -e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8007" title="D4 | Using Laplace transforms to solve IVPs | ver. 8007"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3" alt="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By%7D%20=%202%20%5C,%20%7By''%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="-2 \, {y} = 2 \, {y''} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" title="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20%5Csin%5Cleft(t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" title="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5260" title="D4 | Using Laplace transforms to solve IVPs | ver. 5260"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0" alt="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0" title="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0" data-latex="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20+%2081%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%2027%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%20-3%20,%20y'(0)=%200" alt="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0" title="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0" data-latex="-3 \, {y''} + 81 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -3 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s}{s^{2} + 9} + \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{3 \, s}{s^{2} + 9} + \frac{3 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 3 \, \cos\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7885" title="D4 | Using Laplace transforms to solve IVPs | ver. 7885"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32" alt="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32" title="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32" data-latex="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?48%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20=%20-72%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2032" alt="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32" title="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32" data-latex="48 \, {y} - 3 \, {y''} = -72 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 32">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%2016%7D%20=%20-%5Cfrac%7B1%7D%7B8%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B8%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}" alt="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}" title="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}" data-latex="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}" alt="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}" title="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B24%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%2016%7D%20+%20%5Cfrac%7B32%7D%7Bs%5E%7B2%7D%20-%2016%7D" alt="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}" title="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}" data-latex="\mathcal{L}\{y\}= \frac{24 \, e^{\left(-s\right)}}{s^{2} - 16} + \frac{32}{s^{2} - 16}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s + 4} + \frac{3 \, e^{\left(-s\right)}}{s - 4} - \frac{4}{s + 4} + \frac{4}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}" title="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 3 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1595" title="D4 | Using Laplace transforms to solve IVPs | ver. 1595"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28" alt="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28" title="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28" data-latex="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-42%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20+%203%20%5C,%20%7By'%7D%20+%2036%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-28" alt="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28" title="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28" data-latex="-42 \, \delta\left(t - 3\right) = -3 \, {y''} + 3 \, {y'} + 36 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -28">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D%20=%20-%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}" alt="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}" title="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}" data-latex="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" alt="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B14%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D%20-%20%5Cfrac%7B28%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D" alt="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}" title="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}" data-latex="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 12} - \frac{28}{s^{2} - s - 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 3} - \frac{4}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7940" title="D4 | Using Laplace transforms to solve IVPs | ver. 7940"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" alt="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20=%20-27%20%5C,%20%7By%7D%20-%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="3 \, {y''} = -27 \, {y} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B18%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4782" title="D4 | Using Laplace transforms to solve IVPs | ver. 4782"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" alt="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" title="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%203%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" title="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-9 \, \mathrm{u}\left(t - 2\right) = 3 \, {y} + 3 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{4 \, s}{s^{2} + 1} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(t\right) - 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5063" title="D4 | Using Laplace transforms to solve IVPs | ver. 5063"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20" alt="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20" title="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20" data-latex="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20+%209%20%5C,%20%7By'%7D%20=%20-15%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2020" alt="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20" title="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20" data-latex="12 \, {y} - 3 \, {y''} + 9 \, {y'} = -15 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 20">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20-%204%7D%20=%20-%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 3 \, s - 4} = -\frac{1}{5 \, {\left(s + 1\right)}} + \frac{1}{5 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}" alt="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}" title="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}" data-latex="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}" alt="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}" title="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20-%204%7D%20+%20%5Cfrac%7B20%7D%7Bs%5E%7B2%7D%20-%203%20%5C,%20s%20-%204%7D" alt="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}" title="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}" data-latex="\mathcal{L}\{y\}= \frac{5 \, e^{\left(-3 \, s\right)}}{s^{2} - 3 \, s - 4} + \frac{20}{s^{2} - 3 \, s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 1} + \frac{e^{\left(-3 \, s\right)}}{s - 4} - \frac{4}{s + 1} + \frac{4}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}" title="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5483" title="D4 | Using Laplace transforms to solve IVPs | ver. 5483"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3" alt="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%202%20%5C,%20%7By''%7D%20+%202%20%5C,%20%7By%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="0 = 2 \, {y''} + 2 \, {y} - 4 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" title="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20%5Csin%5Cleft(t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" title="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -2 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 3 \, \sin\left(t\right) + 2 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6040" title="D4 | Using Laplace transforms to solve IVPs | ver. 6040"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" alt="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20=%208%20%5C,%20%7By%7D%20+%2010%20%5C,%20%7By'%7D%20-%2024%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2012" alt="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="-2 \, {y''} = 8 \, {y} + 10 \, {y'} - 24 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20+%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{12}{s^{2} + 5 \, s + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{4}{s + 4} + \frac{4}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-t\right)} - 4 \, e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4799" title="D4 | Using Laplace transforms to solve IVPs | ver. 4799"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0" alt="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0" title="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%20-2%20%5C,%20%7By''%7D%20-%2018%20%5C,%20%7By%7D%20-%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-1%20,%20y'(0)=%200" alt="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0" title="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="0 = -2 \, {y''} - 18 \, {y} - 54 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -1 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8355" title="D4 | Using Laplace transforms to solve IVPs | ver. 8355"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0" alt="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0" title="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0" data-latex="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By%7D%20+%209%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20=%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%204%20,%20y'(0)=%200" alt="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0" title="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0" data-latex="-3 \, {y} + 9 \, \mathrm{u}\left(t - 3\right) = 3 \, {y''} \hspace{2em} y(0)= 4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{4 \, s}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) + 4 \, \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9551" title="D4 | Using Laplace transforms to solve IVPs | ver. 9551"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-54%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%2027%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-54 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} + 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B18%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} - \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) + \sin\left(3 \, t\right) - 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7255" title="D4 | Using Laplace transforms to solve IVPs | ver. 7255"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12" alt="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12" title="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27%20%5C,%20%7By%7D%20+%2036%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20=%203%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2012" alt="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12" title="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="27 \, {y} + 36 \, \delta\left(t - 2\right) = 3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%209%7D%20=%20-%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}" alt="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%209%7D%20+%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20-%209%7D" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-2 \, s\right)}}{s^{2} - 9} + \frac{12}{s^{2} - 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 3} - \frac{2}{s + 3} + \frac{2}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(3 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1937" title="D4 | Using Laplace transforms to solve IVPs | ver. 1937"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20" alt="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20" title="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20" data-latex="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%20-2%20%5C,%20%7By'%7D%20+%202%20%5C,%20%7By''%7D%20-%2012%20%5C,%20%7By%7D%20+%2020%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2020" alt="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20" title="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20" data-latex="0 = -2 \, {y'} + 2 \, {y''} - 12 \, {y} + 20 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 20">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20=%20-%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}" alt="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" alt="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B10%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20+%20%5Cfrac%7B20%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D" alt="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{10 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{20}{s^{2} - s - 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 2} - \frac{2 \, e^{\left(-s\right)}}{s - 3} - \frac{4}{s + 2} + \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6666" title="D4 | Using Laplace transforms to solve IVPs | ver. 6666"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2" alt="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2" title="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2" data-latex="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-16%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20=%20-2%20%5C,%20%7By''%7D%20-%2016%20%5C,%20%7By%7D%20+%2012%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%202" alt="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2" title="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2" data-latex="-16 \, \delta\left(t - 2\right) = -2 \, {y''} - 16 \, {y} + 12 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}" alt="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}" title="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D%20+%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-2 \, s\right)}}{s^{2} - 6 \, s + 8} + \frac{2}{s^{2} - 6 \, s + 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} + \frac{4 \, e^{\left(-2 \, s\right)}}{s - 4} - \frac{1}{s - 2} + \frac{1}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%208%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}" title="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) + e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7939" title="D4 | Using Laplace transforms to solve IVPs | ver. 7939"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-6%20%5C,%20%7By'%7D%20-%204%20%5C,%20%7By%7D%20+%208%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%202%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-6 \, {y'} - 4 \, {y} + 8 \, \delta\left(t - 3\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" alt="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} + 3 \, s + 2} + \frac{3}{s^{2} + 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} + \frac{4 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3}{s + 2} + \frac{3}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t\right)} - 3 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1977" title="D4 | Using Laplace transforms to solve IVPs | ver. 1977"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" alt="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20=%20-2%20%5C,%20%7By%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="2 \, {y''} = -2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) + 4 \, \sin\left(t\right) - 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4682" title="D4 | Using Laplace transforms to solve IVPs | ver. 4682"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" alt="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20-%208%20%5C,%20%7By%7D%20=%208%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-2 \, {y''} - 8 \, {y} = 8 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)" alt="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{4}{s^{2} + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%204%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{e^{\left(-3 \, s\right)}}{s} + \frac{4}{s^{2} + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(2%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20%5Csin%5Cleft(2%20%5C,%20t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) + 2 \, \sin\left(2 \, t\right) - \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2358" title="D4 | Using Laplace transforms to solve IVPs | ver. 2358"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" alt="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" title="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?8%20%5C,%20%7By%7D%20=%20-2%20%5C,%20%7By''%7D%20+%2024%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-8" alt="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" title="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="8 \, {y} = -2 \, {y''} + 24 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20+%204%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{3 \, e^{\left(-2 \, s\right)}}{s} - \frac{8}{s^{2} + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%20-%204%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Csin%5Cleft(2%20%5C,%20t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(2 \, t\right) + 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9229" title="D4 | Using Laplace transforms to solve IVPs | ver. 9229"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20=%202%20%5C,%20%7By%7D%20-%206%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-2 \, {y''} = 2 \, {y} - 6 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20%5Csin%5Cleft(t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) + 3 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9050" title="D4 | Using Laplace transforms to solve IVPs | ver. 9050"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" alt="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" title="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}" alt="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}" title="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}" data-latex="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20=%20-12%20%5C,%20%7By'%7D%20-%2016%20%5C,%20%7By%7D%20+%204%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-8" alt="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" title="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="2 \, {y''} = -12 \, {y'} - 16 \, {y} + 4 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%206%20%5C,%20s%20+%208%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}" title="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}" data-latex="\frac{1}{s^{2} + 6 \, s + 8} = -\frac{1}{2 \, {\left(s + 4\right)}} + \frac{1}{2 \, {\left(s + 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}" alt="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}" title="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%206%20%5C,%20s%20+%208%7D%20-%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20+%206%20%5C,%20s%20+%208%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s^{2} + 6 \, s + 8} - \frac{8}{s^{2} + 6 \, s + 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 4} + \frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{4}{s + 4} - \frac{4}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%208%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}" title="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-4 \, t + 8\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1849" title="D4 | Using Laplace transforms to solve IVPs | ver. 1849"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" alt="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12%20%5C,%20%7By%7D%20-%2010%20%5C,%20%7By'%7D%20+%202%20%5C,%20%7By''%7D%20=%20-6%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="12 \, {y} - 10 \, {y'} + 2 \, {y''} = -6 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}" alt="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-s\right)}}{s^{2} - 5 \, s + 6} - \frac{3}{s^{2} - 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s - 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} + \frac{3}{s - 2} - \frac{3}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D" alt="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4777" title="D4 | Using Laplace transforms to solve IVPs | ver. 4777"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0" alt="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0" title="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20+%2027%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%2027%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0" title="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-3 \, {y''} + 27 \, \mathrm{u}\left(t - 2\right) = 27 \, {y} \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)" alt="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)" title="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)" data-latex="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B9%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{9 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20+%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)" title="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)" data-latex="{y} = -\cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3669" title="D4 | Using Laplace transforms to solve IVPs | ver. 3669"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8" alt="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8" title="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20+%2024%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%2012%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-8" alt="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8" title="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="-3 \, {y''} + 24 \, \mathrm{u}\left(t - 1\right) = 12 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)" alt="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{8}{s^{2} + 4} + \frac{8 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20+%204%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-s\right)}}{s^{2} + 4} + \frac{2 \, e^{\left(-s\right)}}{s} - \frac{8}{s^{2} + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20%5Csin%5Cleft(2%20%5C,%20t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)" title="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -2 \, \cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 4 \, \sin\left(2 \, t\right) + 2 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2860" title="D4 | Using Laplace transforms to solve IVPs | ver. 2860"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0" alt="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0" title="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0" data-latex="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%20-2%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%203%20,%20y'(0)=%200" alt="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0" title="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0" data-latex="2 \, {y''} + 2 \, \mathrm{u}\left(t - 1\right) = -2 \, {y} \hspace{2em} y(0)= 3 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)" alt="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 1} + \frac{3 \, s}{s^{2} + 1} - \frac{e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(t%20-%201%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) + 3 \, \cos\left(t\right) - \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2056" title="D4 | Using Laplace transforms to solve IVPs | ver. 2056"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" alt="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20=%202%20%5C,%20%7By%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-2%20,%20y'(0)=%200" alt="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="-2 \, {y''} = 2 \, {y} - 4 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1520" title="D4 | Using Laplace transforms to solve IVPs | ver. 1520"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3" alt="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20=%2014%20%5C,%20%7By'%7D%20-%2024%20%5C,%20%7By%7D%20+%206%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="2 \, {y''} = 14 \, {y'} - 24 \, {y} + 6 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}" alt="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}" title="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}" data-latex="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} - 7 \, s + 12} - \frac{3}{s^{2} - 7 \, s + 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{3 \, e^{\left(-2 \, s\right)}}{s - 4} + \frac{3}{s - 3} - \frac{3}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%208%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D" alt="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}" title="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}" data-latex="{y} = 3 \, e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8822" title="D4 | Using Laplace transforms to solve IVPs | ver. 8822"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2" alt="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2" title="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2" data-latex="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20-%208%20%5C,%20%7By'%7D%20-%206%20%5C,%20%7By%7D%20+%2016%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%202" alt="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2" title="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2" data-latex="-2 \, {y''} - 8 \, {y'} - 6 \, {y} + 16 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20+%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{2}{s^{2} + 4 \, s + 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 3} + \frac{1}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4850" title="D4 | Using Laplace transforms to solve IVPs | ver. 4850"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3" alt="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3" title="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3" title="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="2 \, {y} + 2 \, {y''} + 2 \, \mathrm{u}\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" alt="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-3 \, s\right)}}{s} - \frac{3}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" title="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)" data-latex="{y} = \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 3 \, \sin\left(t\right) - \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8733" title="D4 | Using Laplace transforms to solve IVPs | ver. 8733"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" alt="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9%20%5C,%20%7By%7D%20-%206%20%5C,%20%7By'%7D%20=%20-3%20%5C,%20%7By''%7D%20+%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2012" alt="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="-9 \, {y} - 6 \, {y'} = -3 \, {y''} + 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}" alt="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}" title="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20+%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 3} + \frac{12}{s^{2} - 2 \, s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 1} + \frac{e^{\left(-2 \, s\right)}}{s - 3} - \frac{3}{s + 1} + \frac{3}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%20e%5E%7B%5Cleft(-t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}" title="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}" data-latex="{y} = e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1998" title="D4 | Using Laplace transforms to solve IVPs | ver. 1998"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6" alt="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?27%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By''%7D%20=%20-27%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%206" alt="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6" title="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="27 \, {y} + 3 \, {y''} = -27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" alt="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B9%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} + \frac{6}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" title="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)" data-latex="{y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6940" title="D4 | Using Laplace transforms to solve IVPs | ver. 6940"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15" alt="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15" title="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15" data-latex="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?45%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%2018%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20+%203%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2015" alt="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15" title="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15" data-latex="45 \, \delta\left(t - 1\right) = 18 \, {y} - 3 \, {y''} + 3 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 15">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20=%20-%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}" alt="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}" alt="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B15%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20+%20%5Cfrac%7B15%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D" alt="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{15 \, e^{\left(-s\right)}}{s^{2} - s - 6} + \frac{15}{s^{2} - s - 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-s\right)}}{s + 2} - \frac{3 \, e^{\left(-s\right)}}{s - 3} - \frac{3}{s + 2} + \frac{3}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 3 \, e^{\left(3 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9595" title="D4 | Using Laplace transforms to solve IVPs | ver. 9595"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20+%2072%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20=%2018%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" title="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-2 \, {y''} + 72 \, \mathrm{u}\left(t - 3\right) = 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B36%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{3}{s^{2} + 9} + \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%209%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + \sin\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3828" title="D4 | Using Laplace transforms to solve IVPs | ver. 3828"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16" alt="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16" title="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16" data-latex="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-9%20%5C,%20%7By%7D%20-%206%20%5C,%20%7By'%7D%20+%203%20%5C,%20%7By''%7D%20=%20-36%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-16" alt="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16" title="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16" data-latex="-9 \, {y} - 6 \, {y'} + 3 \, {y''} = -36 \, \delta\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -16">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" alt="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" alt="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20-%20%5Cfrac%7B16%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{12 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{3 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -3 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2587" title="D4 | Using Laplace transforms to solve IVPs | ver. 2587"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12%20%5C,%20%7By%7D%20-%2015%20%5C,%20%7By'%7D%20=%203%20%5C,%20%7By''%7D%20+%2018%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-12 \, {y} - 15 \, {y'} = 3 \, {y''} + 18 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}" alt="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}" title="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}" data-latex="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} + \frac{3}{s^{2} + 5 \, s + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B1%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 4} - \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{1}{s + 4} + \frac{1}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}" title="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}" data-latex="{y} = -2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) + e^{\left(-t\right)} - e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7867" title="D4 | Using Laplace transforms to solve IVPs | ver. 7867"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4" alt="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20=%20-6%20%5C,%20%7By%7D%20-%208%20%5C,%20%7By'%7D%20+%2016%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="2 \, {y''} = -6 \, {y} - 8 \, {y'} + 16 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D" alt="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{8 \, e^{\left(-s\right)}}{s^{2} + 4 \, s + 3} + \frac{4}{s^{2} + 4 \, s + 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B2%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 3} + \frac{4 \, e^{\left(-s\right)}}{s + 1} - \frac{2}{s + 3} + \frac{2}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5364" title="D4 | Using Laplace transforms to solve IVPs | ver. 5364"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20" alt="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20" title="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20" data-latex="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By'%7D%20+%203%20%5C,%20%7By''%7D%20-%2018%20%5C,%20%7By%7D%20+%2060%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-20" alt="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20" title="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20" data-latex="-3 \, {y'} + 3 \, {y''} - 18 \, {y} + 60 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -20">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20=%20-%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B5%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - s - 6} = -\frac{1}{5 \, {\left(s + 2\right)}} + \frac{1}{5 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}" alt="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" alt="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" title="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B20%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D%20-%20%5Cfrac%7B20%7D%7Bs%5E%7B2%7D%20-%20s%20-%206%7D" alt="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}" title="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}" data-latex="\mathcal{L}\{y\}= -\frac{20 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 6} - \frac{20}{s^{2} - s - 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s + 2} - \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" title="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1464" title="D4 | Using Laplace transforms to solve IVPs | ver. 1464"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6" alt="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6" title="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?12%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%20-2%20%5C,%20%7By''%7D%20+%202%20%5C,%20%7By'%7D%20+%204%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%206" alt="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6" title="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6" data-latex="12 \, \delta\left(t - 1\right) = -2 \, {y''} + 2 \, {y'} + 4 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" alt="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20+%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} + \frac{6}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2}{s + 1} + \frac{2}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8012" title="D4 | Using Laplace transforms to solve IVPs | ver. 8012"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4" alt="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4" title="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20+%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20=%2015%20%5C,%20%7By'%7D%20-%2018%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-4" alt="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4" title="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="3 \, {y''} + 12 \, \delta\left(t - 2\right) = 15 \, {y'} - 18 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" title="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}" data-latex="\frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}" alt="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}" title="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D%20-%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20-%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} - 5 \, s + 6} - \frac{4}{s^{2} - 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{4 \, e^{\left(-2 \, s\right)}}{s - 3} + \frac{4}{s - 2} - \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}" title="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(3 \, t - 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6620" title="D4 | Using Laplace transforms to solve IVPs | ver. 6620"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" alt="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20-%203%20%5C,%20%7By%7D%20=%20-12%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-5%20,%20y'(0)=%200" alt="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="-3 \, {y''} - 3 \, {y} = -12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -5 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{5 \, s}{s^{2} + 1} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%205%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 5 \, \cos\left(t\right) + 4 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3720" title="D4 | Using Laplace transforms to solve IVPs | ver. 3720"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" alt="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-18%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20-%2015%20%5C,%20%7By'%7D%20=%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" title="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-18 \, {y} - 3 \, {y''} - 15 \, {y'} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}" alt="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}" title="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{4}{s^{2} + 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{4 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{4}{s + 3} + \frac{4}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}" title="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -4 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 4 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3928" title="D4 | Using Laplace transforms to solve IVPs | ver. 3928"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" alt="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" title="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20+%209%20%5C,%20%7By%7D%20+%206%20%5C,%20%7By'%7D%20-%2024%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-4" alt="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" title="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="-3 \, {y''} + 9 \, {y} + 6 \, {y'} - 24 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}" alt="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}" alt="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B8%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D%20-%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{8 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 3} - \frac{4}{s^{2} - 2 \, s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} - \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{1}{s + 1} - \frac{1}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - e^{\left(3 \, t\right)} + e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9247" title="D4 | Using Laplace transforms to solve IVPs | ver. 9247"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" alt="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%20-3%20%5C,%20%7By''%7D%20-%203%20%5C,%20%7By%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-2%20,%20y'(0)=%200" alt="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" title="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="0 = -3 \, {y''} - 3 \, {y} + 3 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)" alt="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)" title="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)" data-latex="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{2 \, s}{s^{2} + 1} + \frac{e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20%5Ccos%5Cleft(t%5Cright)%20+%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)" title="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)" data-latex="{y} = -\cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(t\right) + \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4148" title="D4 | Using Laplace transforms to solve IVPs | ver. 4148"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" alt="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%20-18%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By''%7D%20-%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2012" alt="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" title="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12" data-latex="0 = -18 \, {y} - 2 \, {y''} - 54 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= 12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{12}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s} + \frac{12}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 4 \, \sin\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7418" title="D4 | Using Laplace transforms to solve IVPs | ver. 7418"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0" alt="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0" title="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%2054%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%20-2%20,%20y'(0)=%200" alt="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0" title="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0" data-latex="18 \, {y} + 2 \, {y''} + 54 \, \mathrm{u}\left(t - 1\right) = 0 \hspace{2em} y(0)= -2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 3 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4268" title="D4 | Using Laplace transforms to solve IVPs | ver. 4268"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12" alt="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-2%20%5C,%20%7By''%7D%20=%20-4%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By'%7D%20+%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-12" alt="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-2 \, {y''} = -4 \, {y} - 2 \, {y'} + 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}" alt="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20-%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-s\right)}}{s^{2} - s - 2} - \frac{12}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 2} + \frac{4}{s + 1} - \frac{4}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}" title="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}" data-latex="{y} = -2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6975" title="D4 | Using Laplace transforms to solve IVPs | ver. 6975"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" alt="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-18%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20=%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-6" alt="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="-18 \, {y} + 2 \, {y''} = 12 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%209%7D%20=%20-%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 9} = -\frac{1}{6 \, {\left(s + 3\right)}} + \frac{1}{6 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}" alt="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}" title="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}" data-latex="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%209%7D%20-%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20-%209%7D" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-s\right)}}{s^{2} - 9} - \frac{6}{s^{2} - 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B1%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s + 3} + \frac{e^{\left(-s\right)}}{s - 3} + \frac{1}{s + 3} - \frac{1}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}" title="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}" data-latex="{y} = e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t\right)} + e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7421" title="D4 | Using Laplace transforms to solve IVPs | ver. 7421"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4" alt="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4" title="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20=%20-2%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-4" alt="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4" title="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="2 \, {y} + 2 \, {y''} = -2 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)" alt="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)" title="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)" data-latex="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Bs%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 1} - \frac{e^{\left(-2 \, s\right)}}{s} - \frac{4}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20%5Ccos%5Cleft(t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)" title="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)" data-latex="{y} = \cos\left(t - 2\right) \mathrm{u}\left(t - 2\right) - 4 \, \sin\left(t\right) - \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-0362" title="D4 | Using Laplace transforms to solve IVPs | ver. 0362"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14" alt="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14" title="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14" data-latex="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20-%2024%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By'%7D%20-%2028%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-14" alt="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14" title="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14" data-latex="2 \, {y''} - 24 \, {y} - 2 \, {y'} - 28 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -14">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D%20=%20-%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B7%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - s - 12} = -\frac{1}{7 \, {\left(s + 3\right)}} + \frac{1}{7 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}" alt="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}" title="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}" data-latex="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}" alt="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B14%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D%20-%20%5Cfrac%7B14%7D%7Bs%5E%7B2%7D%20-%20s%20-%2012%7D" alt="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}" title="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}" data-latex="\mathcal{L}\{y\}= \frac{14 \, e^{\left(-s\right)}}{s^{2} - s - 12} - \frac{14}{s^{2} - s - 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B2%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 3} + \frac{2 \, e^{\left(-s\right)}}{s - 4} + \frac{2}{s + 3} - \frac{2}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}" title="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1660" title="D4 | Using Laplace transforms to solve IVPs | ver. 1660"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12" alt="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20=%2027%20%5C,%20%7By%7D%20-%2081%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-12" alt="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-3 \, {y''} = 27 \, {y} - 81 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B27%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{12}{s^{2} + 9} + \frac{27 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{12}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%209%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" title="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -3 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) - 4 \, \sin\left(3 \, t\right) + 3 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1646" title="D4 | Using Laplace transforms to solve IVPs | ver. 1646"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4" alt="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4" title="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%202%20%5C,%20%7By''%7D%20-%2012%20%5C,%20%7By'%7D%20+%2016%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%204" alt="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4" title="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4" data-latex="-8 \, \delta\left(t - 1\right) = 2 \, {y''} - 12 \, {y'} + 16 \, {y} \hspace{2em} y(0)= 0 , y'(0)= 4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 6 \, s + 8} = -\frac{1}{2 \, {\left(s - 2\right)}} + \frac{1}{2 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}" alt="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}" title="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}" data-latex="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D%20+%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20-%206%20%5C,%20s%20+%208%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} - 6 \, s + 8} + \frac{4}{s^{2} - 6 \, s + 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s - 2} - \frac{2 \, e^{\left(-s\right)}}{s - 4} - \frac{2}{s - 2} + \frac{2}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D" alt="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}" title="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}" data-latex="{y} = -2 \, e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(2 \, t - 2\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1924" title="D4 | Using Laplace transforms to solve IVPs | ver. 1924"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8" alt="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8" title="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}" alt="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?8%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%206%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20-%208%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-8" alt="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8" title="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8" data-latex="8 \, \delta\left(t - 3\right) = 6 \, {y} + 2 \, {y''} - 8 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -8">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%204%20%5C,%20s%20+%203%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20-%203%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}" title="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}" data-latex="\frac{1}{s^{2} - 4 \, s + 3} = -\frac{1}{2 \, {\left(s - 1\right)}} + \frac{1}{2 \, {\left(s - 3\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}" alt="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}" title="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}" data-latex="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%204%20%5C,%20s%20+%203%7D%20-%20%5Cfrac%7B8%7D%7Bs%5E%7B2%7D%20-%204%20%5C,%20s%20+%203%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 4 \, s + 3} - \frac{8}{s^{2} - 4 \, s + 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%201%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4}{s - 1} - \frac{4}{s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7Bt%7D" alt="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}" title="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}" data-latex="{y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(t - 3\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{t}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7895" title="D4 | Using Laplace transforms to solve IVPs | ver. 7895"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24" alt="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24" title="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24" data-latex="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-24%20%5C,%20%7By%7D%20-%206%20%5C,%20%7By'%7D%20+%2018%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-24" alt="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24" title="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24" data-latex="-24 \, {y} - 6 \, {y'} + 18 \, \delta\left(t - 3\right) = -3 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -24">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20=%20-%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B6%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}" title="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" alt="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" title="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D%20-%20%5Cfrac%7B24%7D%7Bs%5E%7B2%7D%20-%202%20%5C,%20s%20-%208%7D" alt="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}" title="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}" data-latex="\mathcal{L}\{y\}= -\frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{24}{s^{2} - 2 \, s - 8}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}" title="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-3 \, s\right)}}{s + 2} - \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{4}{s + 2} - \frac{4}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" title="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} + 4 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1858" title="D4 | Using Laplace transforms to solve IVPs | ver. 1858"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0" alt="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0" title="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0" data-latex="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%203%20%5C,%20%7By''%7D%20+%2027%20%5C,%20%7By%7D%20-%20108%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%202%20,%20y'(0)=%200" alt="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0" title="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0" data-latex="0 = 3 \, {y''} + 27 \, {y} - 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)" alt="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B36%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} + \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} + \frac{4 \, e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) + 2 \, \cos\left(3 \, t\right) + 4 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-8816" title="D4 | Using Laplace transforms to solve IVPs | ver. 8816"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16" alt="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16" title="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16" data-latex="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-32%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%20-8%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2016" alt="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16" title="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16" data-latex="-32 \, \delta\left(t - 3\right) = -8 \, {y} + 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= 16">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%204%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%202%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - 4} = -\frac{1}{4 \, {\left(s + 2\right)}} + \frac{1}{4 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}" alt="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}" title="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" alt="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" title="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B16%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%204%7D%20+%20%5Cfrac%7B16%7D%7Bs%5E%7B2%7D%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}" title="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}" data-latex="\mathcal{L}\{y\}= -\frac{16 \, e^{\left(-3 \, s\right)}}{s^{2} - 4} + \frac{16}{s^{2} - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 2} - \frac{4}{s + 2} + \frac{4}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" title="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}" data-latex="{y} = -4 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6557" title="D4 | Using Laplace transforms to solve IVPs | ver. 6557"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By%7D%20-%203%20%5C,%20%7By''%7D%20=%2012%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y} - 3 \, {y''} = 12 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{2}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20%5Ccos%5Cleft(t%20-%201%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 4 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3945" title="D4 | Using Laplace transforms to solve IVPs | ver. 3945"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20=%20-3%20%5C,%20%7By'%7D%20-%206%20%5C,%20%7By%7D%20-%2018%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3" title="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-3 \, {y''} = -3 \, {y'} - 6 \, {y} - 18 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}" alt="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}" title="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}" data-latex="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - s - 2} + \frac{3}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-2 \, s\right)}}{s - 2} - \frac{1}{s + 1} + \frac{1}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}" title="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}" data-latex="{y} = 2 \, e^{\left(2 \, t - 4\right)} \mathrm{u}\left(t - 2\right) - 2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(2 \, t\right)} - e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3163" title="D4 | Using Laplace transforms to solve IVPs | ver. 3163"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0" alt="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0" title="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0" data-latex="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?108%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20=%20-3%20%5C,%20%7By''%7D%20-%2027%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%202%20,%20y'(0)=%200" alt="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0" title="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0" data-latex="108 \, \mathrm{u}\left(t - 3\right) = -3 \, {y''} - 27 \, {y} \hspace{2em} y(0)= 2 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)" alt="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B36%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= \frac{2 \, s}{s^{2} + 9} - \frac{36 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B2%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 9} + \frac{2 \, s}{s^{2} + 9} - \frac{4 \, e^{\left(-3 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%209%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 4 \, \cos\left(3 \, t - 9\right) \mathrm{u}\left(t - 3\right) + 2 \, \cos\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2119" title="D4 | Using Laplace transforms to solve IVPs | ver. 2119"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" alt="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" title="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20+%2012%20%5C,%20%7By%7D%20+%2036%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" title="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="3 \, {y''} + 12 \, {y} + 36 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" alt="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} - \frac{12 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} - \frac{3 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%20-%204%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%5Cright)%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" title="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = 3 \, \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(2 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-6871" title="D4 | Using Laplace transforms to solve IVPs | ver. 6871"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20=%203%20%5C,%20%7By%7D%20+%2012%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="-3 \, {y''} = 3 \, {y} + 12 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)" alt="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{2}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 1} - \frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s^{2} + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20%5Ccos%5Cleft(t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20%5Csin%5Cleft(t%5Cright)%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = 4 \, \cos\left(t - 3\right) \mathrm{u}\left(t - 3\right) - 2 \, \sin\left(t\right) - 4 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9450" title="D4 | Using Laplace transforms to solve IVPs | ver. 9450"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" alt="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%2027%20%5C,%20%7By%7D%20+%203%20%5C,%20%7By''%7D%20+%20108%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-3" alt="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" title="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3" data-latex="0 = 27 \, {y} + 3 \, {y''} + 108 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)" alt="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B36%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{3}{s^{2} + 9} - \frac{36 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%209%7D" alt="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" title="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}" data-latex="\mathcal{L}\{y\}= \frac{4 \, s e^{\left(-s\right)}}{s^{2} + 9} - \frac{4 \, e^{\left(-s\right)}}{s} - \frac{3}{s^{2} + 9}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%203%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20%5Csin%5Cleft(3%20%5C,%20t%5Cright)%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)" title="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = 4 \, \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - \sin\left(3 \, t\right) - 4 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-0115" title="D4 | Using Laplace transforms to solve IVPs | ver. 0115"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9" alt="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9" title="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9" data-latex="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20-%2018%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%203%20%5C,%20%7By'%7D%20+%206%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-9" alt="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9" title="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9" data-latex="3 \, {y''} - 18 \, \delta\left(t - 3\right) = 3 \, {y'} + 6 \, {y} \hspace{2em} y(0)= 0 , y'(0)= -9">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20-%202%5Cright)%7D%7D" alt="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" title="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}" data-latex="\frac{1}{s^{2} - s - 2} = -\frac{1}{3 \, {\left(s + 1\right)}} + \frac{1}{3 \, {\left(s - 2\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" alt="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" title="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" data-latex="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B6%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D%20-%20%5Cfrac%7B9%7D%7Bs%5E%7B2%7D%20-%20s%20-%202%7D" alt="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" title="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}" data-latex="\mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - s - 2} - \frac{9}{s^{2} - s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%202%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%201%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s + 1} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{3}{s + 1} - \frac{3}{s - 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%20-%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D" alt="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" title="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}" data-latex="{y} = 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(-t + 3\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-7444" title="D4 | Using Laplace transforms to solve IVPs | ver. 7444"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" alt="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20=%20-36%20%5C,%20%7By%7D%20+%2021%20%5C,%20%7By'%7D%20+%203%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-2" alt="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" title="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2" data-latex="3 \, {y''} = -36 \, {y} + 21 \, {y'} + 3 \, \delta\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -2">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20-%204%7D" alt="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" title="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}" data-latex="\frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}" alt="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}" title="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D%20-%20%5Cfrac%7B2%7D%7Bs%5E%7B2%7D%20-%207%20%5C,%20s%20+%2012%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}" title="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-s\right)}}{s^{2} - 7 \, s + 12} - \frac{2}{s^{2} - 7 \, s + 12}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%203%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20-%204%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}" title="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= -\frac{e^{\left(-s\right)}}{s - 3} + \frac{e^{\left(-s\right)}}{s - 4} + \frac{2}{s - 3} - \frac{2}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20e%5E%7B%5Cleft(3%20%5C,%20t%20-%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D" alt="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}" title="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}" data-latex="{y} = e^{\left(4 \, t - 4\right)} \mathrm{u}\left(t - 1\right) - e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, e^{\left(3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2391" title="D4 | Using Laplace transforms to solve IVPs | ver. 2391"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12" alt="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}" alt="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}" title="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}" data-latex="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?0%20=%209%20%5C,%20%7By%7D%20-%206%20%5C,%20%7By'%7D%20-%203%20%5C,%20%7By''%7D%20-%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-12" alt="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12" title="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="0 = 9 \, {y} - 6 \, {y'} - 3 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%202%20%5C,%20s%20-%203%7D%20=%20-%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20%7B%5Cleft(s%20-%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}" title="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}" data-latex="\frac{1}{s^{2} + 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 3\right)}} + \frac{1}{4 \, {\left(s - 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}" alt="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}" title="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}" alt="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}" title="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}" data-latex="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%202%20%5C,%20s%20-%203%7D%20-%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%202%20%5C,%20s%20-%203%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 2 \, s - 3} - \frac{12}{s^{2} + 2 \, s - 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7Be%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%201%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%201%7D" alt="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}" title="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}" data-latex="\mathcal{L}\{y\}= \frac{e^{\left(-2 \, s\right)}}{s + 3} - \frac{e^{\left(-2 \, s\right)}}{s - 1} + \frac{3}{s + 3} - \frac{3}{s - 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-e%5E%7B%5Cleft(t%20-%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7Bt%7D" alt="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}" title="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}" data-latex="{y} = -e^{\left(t - 2\right)} \mathrm{u}\left(t - 2\right) + e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t\right)} - 3 \, e^{t}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-2359" title="D4 | Using Laplace transforms to solve IVPs | ver. 2359"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0" alt="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20+%208%20%5C,%20%7By%7D%20=%208%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-5%20,%20y'(0)=%200" alt="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0" title="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0" data-latex="2 \, {y''} + 8 \, {y} = 8 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= -5 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)" alt="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)" title="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)" data-latex="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{5 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B5%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7Be%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{s e^{\left(-s\right)}}{s^{2} + 4} - \frac{5 \, s}{s^{2} + 4} + \frac{e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-%5Ccos%5Cleft(2%20%5C,%20t%20-%202%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%205%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%5Cright)%20+%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)" title="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)" data-latex="{y} = -\cos\left(2 \, t - 2\right) \mathrm{u}\left(t - 1\right) - 5 \, \cos\left(2 \, t\right) + \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5570" title="D4 | Using Laplace transforms to solve IVPs | ver. 5570"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12" alt="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12" title="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-10%20%5C,%20%7By'%7D%20-%208%20%5C,%20%7By%7D%20+%2024%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%202%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-12" alt="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12" title="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12" data-latex="-10 \, {y'} - 8 \, {y} + 24 \, \delta\left(t - 1\right) = 2 \, {y''} \hspace{2em} y(0)= 0 , y'(0)= -12">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20=%20-%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B3%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 5 \, s + 4} = -\frac{1}{3 \, {\left(s + 4\right)}} + \frac{1}{3 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B12%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D%20-%20%5Cfrac%7B12%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%204%7D" alt="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}" title="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}" data-latex="\mathcal{L}\{y\}= \frac{12 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 4} - \frac{12}{s^{2} + 5 \, s + 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s + 4} + \frac{4 \, e^{\left(-s\right)}}{s + 1} + \frac{4}{s + 4} - \frac{4}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%204%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}" title="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}" data-latex="{y} = 4 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-4 \, t + 4\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-t\right)} + 4 \, e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-0210" title="D4 | Using Laplace transforms to solve IVPs | ver. 0210"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1" alt="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1" title="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1" data-latex="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?4%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%204%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%206%20%5C,%20%7By'%7D%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-1" alt="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1" title="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1" data-latex="4 \, \delta\left(t - 1\right) = 4 \, {y} + 2 \, {y''} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -1">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" title="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}" data-latex="\frac{1}{s^{2} + 3 \, s + 2} = -\frac{1}{s + 2} + \frac{1}{s + 1}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}" alt="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}" title="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}" data-latex="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%203%20%5C,%20s%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s^{2} + 3 \, s + 2} - \frac{1}{s^{2} + 3 \, s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B1%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}" title="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, e^{\left(-s\right)}}{s + 2} + \frac{2 \, e^{\left(-s\right)}}{s + 1} + \frac{1}{s + 2} - \frac{1}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%202%20%5C,%20e%5E%7B%5Cleft(-t%20+%201%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20e%5E%7B%5Cleft(-t%5Cright)%7D%20+%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D" alt="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}" title="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}" data-latex="{y} = 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 2 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) - e^{\left(-t\right)} + e^{\left(-2 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1842" title="D4 | Using Laplace transforms to solve IVPs | ver. 1842"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0" alt="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0" title="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-3%20%5C,%20%7By''%7D%20+%209%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20=%203%20%5C,%20%7By%7D%20%5Chspace%7B2em%7D%20y(0)=%20-1%20,%20y'(0)=%200" alt="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0" title="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0" data-latex="-3 \, {y''} + 9 \, \mathrm{u}\left(t - 1\right) = 3 \, {y} \hspace{2em} y(0)= -1 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%20s%7D%20=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B1%7D%7Bs%7D" alt="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" title="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}" data-latex="\frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)" alt="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%201%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" title="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{{\left(s^{2} + 1\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20s%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%201%7D%20-%20%5Cfrac%7Bs%7D%7Bs%5E%7B2%7D%20+%201%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, s e^{\left(-s\right)}}{s^{2} + 1} - \frac{s}{s^{2} + 1} + \frac{3 \, e^{\left(-s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20%5Ccos%5Cleft(t%20-%201%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%20%5Ccos%5Cleft(t%5Cright)%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)" title="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="{y} = -3 \, \cos\left(t - 1\right) \mathrm{u}\left(t - 1\right) - \cos\left(t\right) + 3 \, \mathrm{u}\left(t - 1\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1965" title="D4 | Using Laplace transforms to solve IVPs | ver. 1965"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" alt="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" title="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?18%20%5C,%20%7By%7D%20-%2036%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20=%20-2%20%5C,%20%7By''%7D%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" title="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="18 \, {y} - 36 \, \mathrm{u}\left(t - 2\right) = -2 \, {y''} \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%209%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B9%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B9%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" title="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}" data-latex="\frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)" alt="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B18%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%209%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 9} + \frac{18 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B2%20%5C,%20s%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%209%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%209%7D%20+%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{2 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{4 \, s}{s^{2} + 9} + \frac{2 \, e^{\left(-2 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(3%20%5C,%20t%5Cright)%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)" title="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="{y} = -2 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 4 \, \cos\left(3 \, t\right) + 2 \, \mathrm{u}\left(t - 2\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3954" title="D4 | Using Laplace transforms to solve IVPs | ver. 3954"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" alt="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By''%7D%20-%2010%20%5C,%20%7By'%7D%20+%206%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="-12 \, {y} - 2 \, {y''} - 10 \, {y'} + 6 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" alt="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" title="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-3 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-3 \, s\right)}}{s + 3} + \frac{3 \, e^{\left(-3 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%209%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" title="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = 3 \, e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - 3 \, e^{\left(-3 \, t + 9\right)} \mathrm{u}\left(t - 3\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-5810" title="D4 | Using Laplace transforms to solve IVPs | ver. 5810"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0" alt="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0" title="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-8%20%5C,%20%7By%7D%20-%202%20%5C,%20%7By''%7D%20=%20-32%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%20-4%20,%20y'(0)=%200" alt="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0" title="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0" data-latex="-8 \, {y} - 2 \, {y''} = -32 \, \mathrm{u}\left(t - 3\right) \hspace{2em} y(0)= -4 , y'(0)= 0">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B3%7D%20+%204%20%5C,%20s%7D%20=%20-%5Cfrac%7Bs%7D%7B4%20%5C,%20%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B4%20%5C,%20s%7D" alt="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" title="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}" data-latex="\frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" alt="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B16%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7B%7B%5Cleft(s%5E%7B2%7D%20+%204%5Cright)%7D%20s%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" title="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s}{s^{2} + 4} + \frac{16 \, e^{\left(-3 \, s\right)}}{{\left(s^{2} + 4\right)} s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20s%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%7D%20-%20%5Cfrac%7B4%20%5C,%20s%7D%7Bs%5E%7B2%7D%20+%204%7D%20+%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" title="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, s e^{\left(-3 \, s\right)}}{s^{2} + 4} - \frac{4 \, s}{s^{2} + 4} + \frac{4 \, e^{\left(-3 \, s\right)}}{s}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%20-%206%5Cright)%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20%5Ccos%5Cleft(2%20%5C,%20t%5Cright)%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" title="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="{y} = -4 \, \cos\left(2 \, t - 6\right) \mathrm{u}\left(t - 3\right) - 4 \, \cos\left(2 \, t\right) + 4 \, \mathrm{u}\left(t - 3\right)">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-9205" title="D4 | Using Laplace transforms to solve IVPs | ver. 9205"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16" alt="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16" title="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16" data-latex="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" alt="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-32%20%5C,%20%7By%7D%20+%202%20%5C,%20%7By''%7D%20+%2064%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%2016" alt="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16" title="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16" data-latex="-32 \, {y} + 2 \, {y''} + 64 \, \delta\left(t - 3\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 16">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20-%2016%7D%20=%20-%5Cfrac%7B1%7D%7B8%20%5C,%20%7B%5Cleft(s%20+%204%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B8%20%5C,%20%7B%5Cleft(s%20-%204%5Cright)%7D%7D" alt="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" title="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}" data-latex="\frac{1}{s^{2} - 16} = -\frac{1}{8 \, {\left(s + 4\right)}} + \frac{1}{8 \, {\left(s - 4\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}" alt="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}" title="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}" data-latex="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" alt="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" title="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" data-latex="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B32%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20-%2016%7D%20+%20%5Cfrac%7B16%7D%7Bs%5E%7B2%7D%20-%2016%7D" alt="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}" title="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}" data-latex="\mathcal{L}\{y\}= -\frac{32 \, e^{\left(-3 \, s\right)}}{s^{2} - 16} + \frac{16}{s^{2} - 16}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%204%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%20-%204%7D%20-%20%5Cfrac%7B2%7D%7Bs%20+%204%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s + 4} - \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{2}{s + 4} + \frac{2}{s - 4}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%20-%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%20+%2012%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" title="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}" data-latex="{y} = -4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(-4 \, t + 12\right)} \mathrm{u}\left(t - 3\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-4493" title="D4 | Using Laplace transforms to solve IVPs | ver. 4493"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6" alt="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?-12%20%5C,%20%7By'%7D%20-%203%20%5C,%20%7By''%7D%20-%209%20%5C,%20%7By%7D%20=%2012%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-6" alt="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6" title="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6" data-latex="-12 \, {y'} - 3 \, {y''} - 9 \, {y} = 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= -6">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20=%20-%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%203%5Cright)%7D%7D%20+%20%5Cfrac%7B1%7D%7B2%20%5C,%20%7B%5Cleft(s%20+%201%5Cright)%7D%7D" alt="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" title="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}" data-latex="\frac{1}{s^{2} + 4 \, s + 3} = -\frac{1}{2 \, {\left(s + 3\right)}} + \frac{1}{2 \, {\left(s + 1\right)}}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}" alt="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}" title="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D%20-%20%5Cfrac%7B6%7D%7Bs%5E%7B2%7D%20+%204%20%5C,%20s%20+%203%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-2 \, s\right)}}{s^{2} + 4 \, s + 3} - \frac{6}{s^{2} + 4 \, s + 3}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%201%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%201%7D" alt="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}" title="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}" data-latex="\mathcal{L}\{y\}= \frac{2 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{2 \, e^{\left(-2 \, s\right)}}{s + 1} + \frac{3}{s + 3} - \frac{3}{s + 1}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-2%20%5C,%20e%5E%7B%5Cleft(-t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(-t%5Cright)%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}" title="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -2 \, e^{\left(-t + 2\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) - 3 \, e^{\left(-t\right)} + 3 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-3872" title="D4 | Using Laplace transforms to solve IVPs | ver. 3872"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" alt="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" title="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?2%20%5C,%20%7By''%7D%20+%2012%20%5C,%20%7By%7D%20+%2010%20%5C,%20%7By'%7D%20+%208%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%20-4" alt="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" title="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4" data-latex="2 \, {y''} + 12 \, {y} + 10 \, {y'} + 8 \, \delta\left(t - 1\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= -4">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" alt="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" title="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20-%20%5Cfrac%7B4%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{4 \, e^{\left(-s\right)}}{s^{2} + 5 \, s + 6} - \frac{4}{s^{2} + 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%20+%202%7D%20+%20%5Cfrac%7B4%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B4%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}" title="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{4 \, e^{\left(-s\right)}}{s + 3} - \frac{4 \, e^{\left(-s\right)}}{s + 2} + \frac{4}{s + 3} - \frac{4}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%202%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%203%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" title="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -4 \, e^{\left(-2 \, t + 2\right)} \mathrm{u}\left(t - 1\right) + 4 \, e^{\left(-3 \, t + 3\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(-2 \, t\right)} + 4 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item><item ident="D4-1658" title="D4 | Using Laplace transforms to solve IVPs | ver. 1658"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D4.</strong></p><p> Explain how to solve the following IVP. </p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" alt="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3"/></p><p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}"/>.</p></div></mattextxml><mattext texttype="text/html"><div class="exercise-statement">
<p>
<strong>D4.</strong>
</p>
<p> Explain how to solve the following IVP. </p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?3%20%5C,%20%7By''%7D%20+%2018%20%5C,%20%7By%7D%20+%2015%20%5C,%20%7By'%7D%20+%209%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20=%200%20%5Chspace%7B2em%7D%20y(0)=%200%20,%20y'(0)=%203" alt="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" title="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3" data-latex="3 \, {y''} + 18 \, {y} + 15 \, {y'} + 9 \, \delta\left(t - 2\right) = 0 \hspace{2em} y(0)= 0 , y'(0)= 3">
</p>
<p>Hint: <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cfrac%7B1%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20=%20-%5Cfrac%7B1%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B1%7D%7Bs%20+%202%7D" alt="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" title="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}" data-latex="\frac{1}{s^{2} + 5 \, s + 6} = -\frac{1}{s + 3} + \frac{1}{s + 2}">.</p>
</div>
</mattext></material><response_str ident="response1" rcardinality="Single"><render_fib><response_label ident="answer1" rshuffle="No"/></render_fib></response_str></presentation><itemfeedback ident="general_fb"><flow_mat><material><mattextxml><div class="exercise-answer"><h4>Partial Answer:</h4><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}"/></p><p style="text-align:center;"><img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" alt="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" title="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}"/></p></div></mattextxml><mattext texttype="text/html"><div class="exercise-answer">
<h4>Partial Answer:</h4>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D%20+%20%5Cfrac%7B3%7D%7Bs%5E%7B2%7D%20+%205%20%5C,%20s%20+%206%7D" alt="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" title="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}" data-latex="\mathcal{L}\{y\}= -\frac{3 \, e^{\left(-2 \, s\right)}}{s^{2} + 5 \, s + 6} + \frac{3}{s^{2} + 5 \, s + 6}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%5Cmathcal%7BL%7D%5C%7By%5C%7D=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%203%7D%20-%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%20+%202%7D%20-%20%5Cfrac%7B3%7D%7Bs%20+%203%7D%20+%20%5Cfrac%7B3%7D%7Bs%20+%202%7D" alt="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" title="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}" data-latex="\mathcal{L}\{y\}= \frac{3 \, e^{\left(-2 \, s\right)}}{s + 3} - \frac{3 \, e^{\left(-2 \, s\right)}}{s + 2} - \frac{3}{s + 3} + \frac{3}{s + 2}">
</p>
<p style="text-align:center;">
<img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?%7By%7D%20=%20-3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%20+%204%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%20+%206%5Cright)%7D%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20t%5Cright)%7D" alt="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" title="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}" data-latex="{y} = -3 \, e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-3 \, t + 6\right)} \mathrm{u}\left(t - 2\right) + 3 \, e^{\left(-2 \, t\right)} - 3 \, e^{\left(-3 \, t\right)}">
</p>
</div>
</mattext></material></flow_mat></itemfeedback></item></objectbank>
</questestinterop>
