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      <text>19.2.8.2 Solve the equation using the integrating factor</text>
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      <text><![CDATA[<p class="HELM_exercise">Exercise</p>
<p>Using the integrating factors you find, find the general solutions the following equations. Use \(C\) as the arbitary constant.</p>
<ol class="HELM_parts">

    <li>
        <p>\(x \dfrac{\mathrm{d}y}{\mathrm{d}x}+{@a@}xy=xe^{-{@a@}x}\), \(y=\) [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p>
    </li>

    <li>
        <p>\(\dfrac{\mathrm{d}y}{\mathrm{d}x}-(\tan x)y={@b@}\),  \(y=\) [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p>
    </li>

</ol>]]></text>
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      <text><![CDATA[<ol class="HELM_parts">

    <li>
        <p>The standard form is \(\dfrac{\mathrm{d}y}{\mathrm{d}x}+{@a@}y=e^{-{@a@}x}\) for which the integrating factor is \(e^{{@a@}x}\).</p>
<p>Then, on  multiplying through by \(e^{{@a@}x}\) we have
\[e^{{@a@}x}\dfrac{\mathrm{d}y}{\mathrm{d}x}+{@a@}e^{{@a@}x} y=1\]
i.e.\[\dfrac{\mathrm{d}}{\mathrm{d}x}\ (e^{{@a@}x}y)  =1\qquad \text{so that }\qquad e^{{@a@}x}y=x+C\]
leading to\[y=(x+C)e^{-{@a@}x}.\]
 </p>
    </li>

    <li>
        <p>The equation is in standard form and the integrating factor is \(\cos x\).  
Then \[\dfrac{\mathrm{d}}{\mathrm{d}x}(y\cos x )={@b@}\cos x\quad \text{so that }\quad y\cos x=\int {@b@}\cos x\,\mathrm{d}x= {@b@}\sin x+C\]
giving \[ y={@b@}\tan x+C\sec x\]</p>
    </li>

</ol>]]></text>
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      <text>2020120600</text>
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    <questionvariables>
      <text>a: rand(5)+2;
b:rand(5)+2;
ta1: exp(-a*x)*(x+C);
ta2: b*tan(x) + C*sec(x);</text>
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      <text>a)\(x \dfrac{dy}{dx}+{@a@}xy=xe^{-{@a@}x}\),  b) \(\dfrac{dy}{dx}-(\tan x)y={@b@}\)</text>
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