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      <text>20.2.3.3 The first shift theorem</text>
    </name>
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      <text><![CDATA[<p></p>

<p>We can also employ the first shift theorem to determine some inverse Laplace transforms.</p>
<br>
<p class="HELM_exercise">Exercise</p>
<p>Find the inverse Laplace transform of \({F(s)=\displaystyle {3\over {s^2-2s-8}}}\).</p>
<ol class="HELM_parts">
    <li>Begin by completing the square in the denominator:
        <br>
        <p>[[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p>
    </li>
    <li>Recalling that \({\cal L}\{\sinh 3t\ u(t)\}={3\over {s^2-9}}\) (from Task) complete the inversion using the first shift theorem:
        <br>
        <p>\({\cal L}^{-1}\{F(s)\} = \) [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p>
    </li>
</ol>]]></text>
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      <text><![CDATA[<p>Find the inverse Laplace transform of \({F(s)=\displaystyle {3\over {s^2-2s-8}}}\).</p>

<ol class="HELM_parts">
  <li><p>Begin by completing the square in the denominator:</p>
    <p>\(\dfrac{3}{s^2-2s-8} = \dfrac{3}{(s − 1)^2 − 9}\)</p></li>
  
    <li>You should obtain
        <p>\(\qquad {\cal L}^{-1}\left\{\dfrac{3}{(s − 1)^2 − 9}\right\} = \text{e}^{t}\sinh 3t u(t)\)
        </p>
        <p>Here, in the notation of the shift theorem:</p>
        <p>\(\qquad f(t) = \sinh 3t u(t) \qquad F(s) = \dfrac{3}{s^2-9} \qquad\) and \(\quad a = -1 \)</p>
    </li>
</ol>]]></text>
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    <stackversion>
      <text>2023010400</text>
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      <text>ta1: 3/((s-1)^2-9);
ta2: e^t*sinh(3*t)*u(t);</text>
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