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<quiz>
  <question type="stack">
    <name>
      <text>19.3.2.4 Auxiliary Equations</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Write down, but do not solve nor simplify, the auxiliary equations of the following:</p>
<ol class="HELM_parts">
  
  <li><p>\( {@a1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@a2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} + {@a3@}y = 0\).</p>
    <p>The auxiliary equation is [[input:ans1]] [[validation:ans1]] \(= 0\). [[feedback:prt1]]</p></li>
    
    <li><p>\( {@b1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@b2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} - {@b3@}y = 0\).</p>
    <p>The auxiliary equation is [[input:ans2]] [[validation:ans2]] \(= 0\). [[feedback:prt2]]</p></li>
      
      <li><p>\( {@c1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@c3@}y = 0\).</p>
    <p>The auxiliary equation is [[input:ans3]] [[validation:ans3]] \(= 0\). [[feedback:prt3]]</p></li>
  
  <li><p>\( {@d1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@d2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\).</p>
    <p>The auxiliary equation is [[input:ans4]] [[validation:ans4]] \(= 0\). [[feedback:prt4]]</p></li></ol>]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<p> For each part of the question we make use of the substitution \(y= e^{kx}\). With this substitution, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = ke^{kx}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = k^2e^{kx}\).</p>

<ol class="HELM_parts">
  <li><p>We are given the equation \[\begin{aligned} {@a1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@a2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} + {@a3@}y &amp;= 0, \\
    \implies {@a1@}k^2e^{kx} + {@a2@}ke^{kx} + {@a3@}e^{kx} &amp;= 0, \\
    \implies {@ta1@} &amp;= 0. \end{aligned}\]</p></li>
  
  <li><p>We are given the equation \[\begin{aligned} {@b1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@b2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} - {@b3@}y &amp;= 0, \\
    \implies {@b1@}k^2e^{kx} + {@b2@}ke^{kx} - {@b3@}e^{kx} &amp;= 0, \\
    \implies {@ta2@} &amp;= 0. \end{aligned}\]</p></li>
  
  <li><p>We are given the equation \[\begin{aligned} {@c1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@c3@}y &amp;= 0, \\
    \implies {@c1@}k^2e^{kx} + {@c3@}e^{kx} &amp;= 0, \\
    \implies {@ta3@} &amp;= 0. \end{aligned}\]</p></li>
  
  <li><p>We are given the equation \[\begin{aligned} {@d1@}\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + {@d2@}\dfrac{\mathrm{d}y}{\mathrm{d}x} &amp;= 0, \\
    \implies {@d1@}k^2e^{kx} + {@d2@}ke^{kx} &amp;= 0, \\
    \implies {@ta4@} &amp;= 0. \end{aligned}\]</p></li>
  
</ol>]]></text>
    </generalfeedback>
    <defaultgrade>1</defaultgrade>
    <penalty>0.1</penalty>
    <hidden>0</hidden>
    <idnumber/>
    <stackversion>
      <text>2020120600</text>
    </stackversion>
    <questionvariables>
      <text>a1: 2+rand(3);
a2: 2+rand(3);
a3: 2+rand(3);

ta1: a1*k^2 + a2*k + a3;

b1: 2+ rand(3);
b2: 6+rand(3);
b3: 2 + rand(3);

ta2: b1*k^2 + b2*k - b3;

c1: 3+rand(3);
c3: 6+rand(3);

ta3: c1*k^2 + c3;

d1: 2 + rand(3);
d2: 2 + rand(3);

ta4: d1*k^2 + d2*k;</text>
    </questionvariables>
    <specificfeedback format="html">
      <text/>
    </specificfeedback>
    <questionnote format="moodle_auto_format">
      <text>\( {@ta1@} = 0, {@ta2@} = 0, {@ta3@} = 0, {@ta4@} = 0\)</text>
    </questionnote>
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      <text/>
    </questiondescription>
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    <prtcorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
    </prtcorrect>
    <prtpartiallycorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:orange;"><i class="fa fa-adjust"></i></span> Your answer is partially correct.]]></text>
    </prtpartiallycorrect>
    <prtincorrect format="html">
      <text><![CDATA[<span style="font-size: 1.5em; color:red;"><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
    </prtincorrect>
    <decimals>.</decimals>
    <scientificnotation>*10</scientificnotation>
    <multiplicationsign>none</multiplicationsign>
    <sqrtsign>1</sqrtsign>
    <complexno>i</complexno>
    <inversetrig>cos-1</inversetrig>
    <logicsymbol>lang</logicsymbol>
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