0007.qti

<item ident="D2-5950" title="D2 | Laplace transforms from formula and definition | ver. 5950">
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  <presentation>
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        <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>
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      <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;

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    <response_str ident="response1" rcardinality="Single">
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        <response_label ident="answer1" rshuffle="No"/>
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  <itemfeedback ident="general_fb">
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            <h4>Partial Answer:</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>
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        <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;

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