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      <text>19.2.7.1 Finding the integrating factor for linear  ODEs</text>
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      <text><![CDATA[<h4>Finding the integrating factor for linear  ODEs</h4>
<p> The differential equation governing the current \(i\) in a circuit with inductance \(L\) and resistance \(R\) in series subject to a constant applied electromotive force \(E\cos\omega t\), where \(E\) and \(\omega\) are constants, is
  \[ L \dfrac{\mathrm{d}i}{\mathrm{d}t}+Ri=E\cos \omega t. \quad \quad (1)\]&nbsp;</p>
<p>This is an example of a linear differential equation in which \(i\) is the dependent variable and \(t\) is the independent variable. The general standard form of a linear first order differential equation is normally written with '\(y\)' as the dependent variable and with '\(x\)' as the independent variable and arranged so that the coefficient of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) is \(1\). That is, it takes the form:
\[  \dfrac{\mathrm{d}y}{\mathrm{d}x}+f(x)y=g(x) \quad \quad (2)\]&nbsp; in which \(f(x)\) and \(g(x)\) are functions of \(x\). </p>
<p> Comparing (1) and (2), \(x\) is replaced by \(t\) and \(y\) by \(i\) to produce \[\dfrac{\mathrm{d}i}{\mathrm{d}t}+f(t)\ i=g(t).\] The function \(f(t)\) is the coefficient of the dependent variable in the differential equation.  We shall describe the method of finding the
integrating factor for (1) and then generalise it to a linear differential equation written in standard form.</p>

<p><strong> Step 1</strong> Write the differential equation in standard form i.e. with the coefficient of the derivative equal to \(1\). Here we need to divide through by \(L\):
  \[ \dfrac{\mathrm{d}i}{\mathrm{d}t}+\dfrac{R}{L}i= \dfrac{E}{L}\cos \omega t.\]</p>

<p><strong> Step 2</strong>  Integrate the coefficient of the dependent variable  (that is, \(f(t)= R/L\)) with respect to the independent variable (that is, \(t\)), and ignoring the constant of integration
  \[ \int \dfrac{R}{L}\, \mathrm{d}t= \dfrac{R}{L} t.\]</p>

<p><strong> Step 3</strong>\quad Take the exponential of the function obtained in Step 2. </p> 
<p>This is the integrating factor (I.F.)
\[ \text{I.F.} =\text{e}^{Rt/L}.\]</p><p>This leads to the following Key Point on integrating factors:</p>
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<h4>Key Point</h4>
<p>The linear differential equation (written in standard form):
\[\dfrac{\mathrm{d}y}{\mathrm{d}x}+f(x)y=g(x)\qquad \text{has an integrating factor}\qquad
\text{I.F.} = \text{exp}\left[\int f(x)\mathrm{d}x\right]\] .</p>
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