CoCalc -- D2.qti

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<?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="D2">
    <qtimetadata>
      <qtimetadatafield><fieldlabel>bank_title</fieldlabel><fieldentry>Differential Equations -- D2</fieldentry></qtimetadatafield>
    </qtimetadata>
  <item ident="D2-9069" title="D2 | Laplace transforms from formula and definition | ver. 9069"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" alt="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} + 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6021" title="D2 | Laplace transforms from formula and definition | ver. 6021"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" alt="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 5\right) + 3 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 3} - 3 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1783" title="D2 | Laplace transforms from formula and definition | ver. 1783"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" alt="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -2 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5611" title="D2 | Laplace transforms from formula and definition | ver. 5611"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 1\right) + 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8145" title="D2 | Laplace transforms from formula and definition | ver. 8145"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" alt="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 3} - 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9392" title="D2 | Laplace transforms from formula and definition | ver. 9392"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" alt="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 2 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3244" title="D2 | Laplace transforms from formula and definition | ver. 3244"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -2 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 2} - 2 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5950" title="D2 | Laplace transforms from formula and definition | ver. 5950"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} - 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6682" title="D2 | Laplace transforms from formula and definition | ver. 6682"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" alt="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 4} + 2 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1464" title="D2 | Laplace transforms from formula and definition | 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>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-7468" title="D2 | Laplace transforms from formula and definition | ver. 7468"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" alt="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" title="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" title="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9969" title="D2 | Laplace transforms from formula and definition | ver. 9969"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" alt="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 3\right) + 4 \, e^{\left(2 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 2} - 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2519" title="D2 | Laplace transforms from formula and definition | ver. 2519"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 3\right) - 4 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 4} - 2 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9688" title="D2 | Laplace transforms from formula and definition | ver. 9688"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -3 \, \delta\left(t - 5\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1736" title="D2 | Laplace transforms from formula and definition | ver. 1736"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" alt="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 1\right) + 5 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9342" title="D2 | Laplace transforms from formula and definition | ver. 9342"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" alt="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 2\right) - 4 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4360" title="D2 | Laplace transforms from formula and definition | ver. 4360"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" alt="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0153" title="D2 | Laplace transforms from formula and definition | ver. 0153"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" alt="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" title="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" title="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -2 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} - 2 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5349" title="D2 | Laplace transforms from formula and definition | ver. 5349"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" alt="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 1\right) + 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-5 \, s\right)}}{s} + \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3230" title="D2 | Laplace transforms from formula and definition | ver. 3230"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)" alt="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} + 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5246" title="D2 | Laplace transforms from formula and definition | ver. 5246"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 4} + 4 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4732" title="D2 | Laplace transforms from formula and definition | ver. 4732"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" alt="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 3\right) + 4 \, e^{\left(5 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{4}{s - 5} + 2 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5400" title="D2 | Laplace transforms from formula and definition | ver. 5400"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" alt="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5874" title="D2 | Laplace transforms from formula and definition | ver. 5874"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -3 \, \delta\left(t - 2\right) + 5 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 3} - 3 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1262" title="D2 | Laplace transforms from formula and definition | ver. 1262"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" alt="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" title="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" title="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 2 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 5} + 2 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8653" title="D2 | Laplace transforms from formula and definition | ver. 8653"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)" alt="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 5\right) - 3 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 2 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2919" title="D2 | Laplace transforms from formula and definition | ver. 2919"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" alt="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) - 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0869" title="D2 | Laplace transforms from formula and definition | ver. 0869"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" alt="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%205%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 5} + 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5419" title="D2 | Laplace transforms from formula and definition | ver. 5419"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 4\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 4 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6659" title="D2 | Laplace transforms from formula and definition | ver. 6659"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" alt="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 4} + 2 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5722" title="D2 | Laplace transforms from formula and definition | ver. 5722"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" alt="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 1\right) - 4 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3702" title="D2 | Laplace transforms from formula and definition | ver. 3702"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" alt="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} - \frac{5}{s - 5} - 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9395" title="D2 | Laplace transforms from formula and definition | ver. 9395"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" alt="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 4\right) + 4 \, e^{\left(5 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 5} - 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3561" title="D2 | Laplace transforms from formula and definition | ver. 3561"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" alt="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" title="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" title="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} + 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1297" title="D2 | Laplace transforms from formula and definition | ver. 1297"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)" alt="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-7476" title="D2 | Laplace transforms from formula and definition | ver. 7476"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" alt="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 4\right) + 2 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-4 \, s\right)}}{s} + \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6255" title="D2 | Laplace transforms from formula and definition | ver. 6255"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" alt="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{5}{s - 4} + 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2174" title="D2 | Laplace transforms from formula and definition | ver. 2174"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 3 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-7884" title="D2 | Laplace transforms from formula and definition | ver. 7884"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0796" title="D2 | Laplace transforms from formula and definition | ver. 0796"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" alt="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" title="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" title="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{5}{s - 4} + 2 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0457" title="D2 | Laplace transforms from formula and definition | ver. 0457"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" alt="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} + \frac{3}{s - 5} + 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3706" title="D2 | Laplace transforms from formula and definition | ver. 3706"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" alt="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8215" title="D2 | Laplace transforms from formula and definition | ver. 8215"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" alt="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 4 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6703" title="D2 | Laplace transforms from formula and definition | ver. 6703"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" alt="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(5 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6768" title="D2 | Laplace transforms from formula and definition | ver. 6768"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" alt="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 5\right) + 2 \, e^{\left(5 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{2}{s - 5} + 4 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6456" title="D2 | Laplace transforms from formula and definition | ver. 6456"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" alt="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" title="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" title="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} + 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9022" title="D2 | Laplace transforms from formula and definition | ver. 9022"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" alt="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 3\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{3}{s - 2} + 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2685" title="D2 | Laplace transforms from formula and definition | ver. 2685"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2605" title="D2 | Laplace transforms from formula and definition | ver. 2605"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" alt="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" title="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 3 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{3 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4354" title="D2 | Laplace transforms from formula and definition | ver. 4354"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 3 \, \delta\left(t - 3\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2995" title="D2 | Laplace transforms from formula and definition | ver. 2995"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 5} - 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6407" title="D2 | Laplace transforms from formula and definition | ver. 6407"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" alt="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 2} + 4 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8958" title="D2 | Laplace transforms from formula and definition | ver. 8958"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)" alt="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 4\right) - 3 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0681" title="D2 | Laplace transforms from formula and definition | ver. 0681"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" alt="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 2 \, \delta\left(t - 3\right) - 3 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{3}{s - 5} + 2 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8608" title="D2 | Laplace transforms from formula and definition | ver. 8608"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)" alt="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)" title="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)" title="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -4 \, \delta\left(t - 5\right) + 3 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%205%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{3}{s - 5} - 4 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8842" title="D2 | Laplace transforms from formula and definition | ver. 8842"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" alt="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" title="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" title="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 4 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 4} - 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3246" title="D2 | Laplace transforms from formula and definition | ver. 3246"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 3 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9094" title="D2 | Laplace transforms from formula and definition | ver. 9094"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" alt="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -2 \, \delta\left(t - 3\right) - 2 \, e^{\left(5 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 5} - 2 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6294" title="D2 | Laplace transforms from formula and definition | ver. 6294"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" alt="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 5 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%203%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 3} + 2 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9579" title="D2 | Laplace transforms from formula and definition | ver. 9579"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" alt="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" title="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" title="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 5 \, \delta\left(t - 4\right) + 2 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} + 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-7925" title="D2 | Laplace transforms from formula and definition | ver. 7925"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" alt="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 3\right) - 5 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%203%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{5}{s - 3} - 5 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4524" title="D2 | Laplace transforms from formula and definition | ver. 4524"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)" alt="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 1\right) - 3 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-s\right)}}{s} - \frac{3}{s - 2} + 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6533" title="D2 | Laplace transforms from formula and definition | ver. 6533"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" alt="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" title="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -4 \, \delta\left(t - 3\right) - 3 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 2} - 4 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6126" title="D2 | Laplace transforms from formula and definition | ver. 6126"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" alt="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 4\right) + 3 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%202%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 2} - 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3674" title="D2 | Laplace transforms from formula and definition | ver. 3674"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 4 \, \delta\left(t - 5\right) + 5 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%205%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 5} + 4 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2478" title="D2 | Laplace transforms from formula and definition | ver. 2478"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" alt="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 4 \, \delta\left(t - 4\right) + 2 \, e^{\left(2 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2217" title="D2 | Laplace transforms from formula and definition | ver. 2217"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)" alt="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} - 2 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1031" title="D2 | Laplace transforms from formula and definition | ver. 1031"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 3\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6104" title="D2 | Laplace transforms from formula and definition | ver. 6104"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" alt="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 3 \, \delta\left(t - 4\right) - 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} - \frac{4}{s - 3} + 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6624" title="D2 | Laplace transforms from formula and definition | ver. 6624"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" alt="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 5\right) - 2 \, e^{\left(5 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 5} - 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-3375" title="D2 | Laplace transforms from formula and definition | ver. 3375"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" alt="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2618" title="D2 | Laplace transforms from formula and definition | ver. 2618"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 3\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 4} - 3 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5537" title="D2 | Laplace transforms from formula and definition | ver. 5537"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" alt="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" title="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" title="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 3 \, \delta\left(t - 5\right) + 2 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 4} + 3 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6765" title="D2 | Laplace transforms from formula and definition | ver. 6765"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" alt="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" title="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" title="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 5 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%202%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-2 \, s\right)}}{s} + \frac{5}{s - 2} + 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4919" title="D2 | Laplace transforms from formula and definition | ver. 4919"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" alt="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 2\right) + 5 \, e^{\left(2 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%202%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0508" title="D2 | Laplace transforms from formula and definition | ver. 0508"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" alt="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" title="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -5 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-2 \, s\right)}}{s} - \frac{3}{s - 3} - 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2574" title="D2 | Laplace transforms from formula and definition | ver. 2574"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" title="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -4 \, \delta\left(t - 4\right) + 2 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 4} - 4 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6248" title="D2 | Laplace transforms from formula and definition | ver. 6248"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)" alt="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)" title="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)" title="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 2 \, \delta\left(t - 4\right) + 4 \, e^{\left(4 \, t\right)} + 5 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%204%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 4} + 2 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2494" title="D2 | Laplace transforms from formula and definition | ver. 2494"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" alt="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 2 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%202%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{2}{s - 2} + 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8129" title="D2 | Laplace transforms from formula and definition | ver. 8129"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -5 \, \delta\left(t - 2\right) - 4 \, e^{\left(5 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 5} - 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1919" title="D2 | Laplace transforms from formula and definition | ver. 1919"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 4 \, \delta\left(t - 4\right) + 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} + \frac{5}{s - 4} + 4 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9136" title="D2 | Laplace transforms from formula and definition | ver. 9136"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -5 \, \delta\left(t - 4\right) - 2 \, e^{\left(3 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 5 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8407" title="D2 | Laplace transforms from formula and definition | ver. 8407"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" alt="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" title="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 4 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-4 \, s\right)}}{s} + \frac{4}{s - 3} + 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4575" title="D2 | Laplace transforms from formula and definition | ver. 4575"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" alt="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = 3 \, \delta\left(t - 1\right) + 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%202%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2755" title="D2 | Laplace transforms from formula and definition | ver. 2755"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" alt="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%205%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 3\right) + 5 \, e^{\left(2 \, t\right)} - 5 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B5%7D%7Bs%20-%202%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-3 \, s\right)}}{s} + \frac{5}{s - 2} + 3 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1805" title="D2 | Laplace transforms from formula and definition | ver. 1805"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" alt="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" title="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)" data-latex="y = 5 \, \delta\left(t - 5\right) - 5 \, e^{\left(4 \, t\right)} - 3 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%204%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 4} + 5 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1733" title="D2 | Laplace transforms from formula and definition | ver. 1733"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" alt="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" title="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%204%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(5%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" title="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 4 \, \delta\left(t - 1\right) + 4 \, e^{\left(5 \, t\right)} + 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%205%7D%20+%204%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{4}{s - 5} + 4 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9316" title="D2 | Laplace transforms from formula and definition | ver. 9316"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" alt="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 5\right) - 4 \, e^{\left(2 \, t\right)} + 2 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%202%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-3 \, s\right)}}{s} - \frac{4}{s - 2} + 3 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-6399" title="D2 | Laplace transforms from formula and definition | ver. 6399"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" alt="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%203%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = 3 \, \delta\left(t - 2\right) - 3 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%203%7D%20+%203%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} - \frac{3}{s - 3} + 3 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9380" title="D2 | Laplace transforms from formula and definition | ver. 9380"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 1\right) - 2 \, e^{\left(3 \, t\right)} + 5 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-5 \, s\right)}}{s} - \frac{2}{s - 3} - 3 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-4307" title="D2 | Laplace transforms from formula and definition | ver. 4307"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" alt="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%202%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" title="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)" data-latex="y = 2 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 2 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20+%202%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{2 \, e^{\left(-2 \, s\right)}}{s} - \frac{4}{s - 3} + 2 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2472" title="D2 | Laplace transforms from formula and definition | ver. 2472"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" alt="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20+%202%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%202%5Cright)" alt="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" title="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)" data-latex="y = -4 \, \delta\left(t - 5\right) + 2 \, e^{\left(3 \, t\right)} - 4 \, \mathrm{u}\left(t - 2\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B2%7D%7Bs%20-%203%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-2 \, s\right)}}{s} + \frac{2}{s - 3} - 4 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-9419" title="D2 | Laplace transforms from formula and definition | ver. 9419"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" alt="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-5%20%5C,%20%5Cdelta%5Cleft(t%20-%201%5Cright)%20-%202%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%202%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" title="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)" data-latex="y = -5 \, \delta\left(t - 1\right) - 2 \, e^{\left(4 \, t\right)} + 2 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B2%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B2%7D%7Bs%20-%204%7D%20-%205%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}" title="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}" data-latex="\mathcal{L}\{y\} = \frac{2 \, e^{\left(-4 \, s\right)}}{s} - \frac{2}{s - 4} - 5 \, e^{\left(-s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-7375" title="D2 | Laplace transforms from formula and definition | ver. 7375"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" alt="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20+%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" title="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -2 \, \delta\left(t - 2\right) + 4 \, e^{\left(3 \, t\right)} - 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-s\right)}}{s} + \frac{4}{s - 3} - 2 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8949" title="D2 | Laplace transforms from formula and definition | ver. 8949"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20+%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 3\right) - 3 \, e^{\left(4 \, t\right)} + 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-8792" title="D2 | Laplace transforms from formula and definition | ver. 8792"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" alt="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-2%20%5C,%20%5Cdelta%5Cleft(t%20-%203%5Cright)%20+%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%203%5Cright)" alt="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" title="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)" data-latex="y = -2 \, \delta\left(t - 3\right) + 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 3\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20+%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20-%202%20%5C,%20e%5E%7B%5Cleft(-3%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-3 \, s\right)}}{s} + \frac{3}{s - 4} - 2 \, e^{\left(-3 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-2595" title="D2 | Laplace transforms from formula and definition | ver. 2595"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%204%5Cright)%20-%203%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%204%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 4\right) - 3 \, e^{\left(4 \, t\right)} - 4 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B4%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B3%7D%7Bs%20-%204%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{4 \, e^{\left(-5 \, s\right)}}{s} - \frac{3}{s - 4} - 3 \, e^{\left(-4 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-1628" title="D2 | Laplace transforms from formula and definition | ver. 1628"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" alt="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%205%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(4%20%5C,%20t%5Cright)%7D%20-%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%204%5Cright)" alt="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" title="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)" data-latex="y = 5 \, \delta\left(t - 2\right) - 4 \, e^{\left(4 \, t\right)} - 5 \, \mathrm{u}\left(t - 4\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-4%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%204%7D%20+%205%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{5 \, e^{\left(-4 \, s\right)}}{s} - \frac{4}{s - 4} + 5 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-5760" title="D2 | Laplace transforms from formula and definition | ver. 5760"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" alt="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-3%20%5C,%20%5Cdelta%5Cleft(t%20-%205%5Cright)%20-%204%20%5C,%20e%5E%7B%5Cleft(3%20%5C,%20t%5Cright)%7D%20-%203%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%205%5Cright)" alt="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" title="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)" data-latex="y = -3 \, \delta\left(t - 5\right) - 4 \, e^{\left(3 \, t\right)} - 3 \, \mathrm{u}\left(t - 5\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20-%5Cfrac%7B3%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B4%7D%7Bs%20-%203%7D%20-%203%20%5C,%20e%5E%7B%5Cleft(-5%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" title="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}" data-latex="\mathcal{L}\{y\} = -\frac{3 \, e^{\left(-5 \, s\right)}}{s} - \frac{4}{s - 3} - 3 \, e^{\left(-5 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item><item ident="D2-0670" title="D2 | Laplace transforms from formula and definition | ver. 0670"><itemmetadata><qtimetadata><qtimetadatafield><fieldlabel>question_type</fieldlabel><fieldentry>essay_question</fieldentry></qtimetadatafield></qtimetadata></itemmetadata><presentation><material><mattextxml><div class="exercise-statement"><p><strong>D2.</strong></p><p> Compute the Laplace transform <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?\mathcal{L}\{y\}" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"/> of <img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" alt="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"/> by using a transform table. </p><p> Then show how the integral definition of the Laplace transform to obtains same result. </p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-statement"&gt;
  &lt;p&gt;
    &lt;strong&gt;D2.&lt;/strong&gt;
  &lt;/p&gt;
  &lt;p&gt; Compute the Laplace transform &lt;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" alt="\mathcal{L}\{y\}" title="\mathcal{L}\{y\}" data-latex="\mathcal{L}\{y\}"&gt; of &lt;img style="border:1px #ddd solid;padding:5px;border-radius:5px;" src="https://latex.codecogs.com/svg.latex?y%20=%20-4%20%5C,%20%5Cdelta%5Cleft(t%20-%202%5Cright)%20-%205%20%5C,%20e%5E%7B%5Cleft(2%20%5C,%20t%5Cright)%7D%20+%205%20%5C,%20%5Cmathrm%7Bu%7D%5Cleft(t%20-%201%5Cright)" alt="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" title="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)" data-latex="y = -4 \, \delta\left(t - 2\right) - 5 \, e^{\left(2 \, t\right)} + 5 \, \mathrm{u}\left(t - 1\right)"&gt; by using a transform table. &lt;/p&gt;
  &lt;p&gt; Then show how the integral definition of the Laplace transform to obtains same result. &lt;/p&gt;
&lt;/div&gt;

</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(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}"/></p></div></mattextxml><mattext texttype="text/html">&lt;div class="exercise-answer"&gt;
  &lt;h4&gt;Partial Answer:&lt;/h4&gt;
  &lt;p style="text-align:center;"&gt;
    &lt;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=%20%5Cfrac%7B5%20%5C,%20e%5E%7B%5Cleft(-s%5Cright)%7D%7D%7Bs%7D%20-%20%5Cfrac%7B5%7D%7Bs%20-%202%7D%20-%204%20%5C,%20e%5E%7B%5Cleft(-2%20%5C,%20s%5Cright)%7D" alt="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" title="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}" data-latex="\mathcal{L}\{y\} = \frac{5 \, e^{\left(-s\right)}}{s} - \frac{5}{s - 2} - 4 \, e^{\left(-2 \, s\right)}"&gt;
  &lt;/p&gt;
&lt;/div&gt;

</mattext></material></flow_mat></itemfeedback></item></objectbank>
</questestinterop>
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