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      <text>19.3.6.1 Inhomogeneous term in the complementary function</text>
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      <text><![CDATA[<h4>Inhomogeneous term in the complementary function</h4>
<p>Occasionally, you will come accross a differential equation \(\displaystyle a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b \frac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x)\) for which the inhomogeneous term, \(f(x)\), forms part of the complementary function. One such example is the equation</p>
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - \frac{\mathrm{d}y}{\mathrm{d}x} - 6y = e^{3x}.\]
<p>It is straightforward to check that the complementary function is \(y_\text{cf} = Ae^{3x} + Be^{-2x}\). Note that the first of these terms has the same form as the inhomogeneous term, \(e^{3x}\), on the right-hand side of the differential equation.</p>
<p>You should verify for yourself that trying a particular integral of the form \(y_p(x) = \alpha e^{3x}\) will not work in a case like this. Can you see why?</p>
<p>Instead, try a particular integral of the form \(y_\text{p}(x) = \alpha x e^{3x}\). Verify that</p>
\[\frac{\mathrm{d}y_\text{p}}{\mathrm{d}x} = \alpha e^{3x}(3x + 1) \qquad \text{and} \qquad \frac{\mathrm{d}^2y_\text{p}}{\mathrm{d}x^2} = \alpha e^{3x}(9x + 6).\]
<p>Substitute these expressions into the differential equation to find \(\displaystyle\alpha = \frac{1}{5}\).</p>
<p>Finally, the particular integral is \(\displaystyle y_\text{p}(x) = \frac{1}{5}xe^{3x}\), and so the general solution to the differential equation is:</p>
\[y = Ae^{3x} + Be^{-2x} + \frac{1}{5}xe^{3x}\]
<p>This shows a generally effective method &ndash; where the inhomogeneous term \(f(x)\) appears in the complementary function use as a trial particular integral \(x\) times what would otherwise be used.</p>

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<h4>Key Point</h4>
<p>When solving</p>
\[a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b \frac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x),\]
<p>if the inhomogeneous \(f(x)\) appears in the complementary function use as a trial particular integral \(x\) times what would otherwise be used.</p>
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