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      <text>19.3.4.3 Particular integral example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>Obtain a particular integral of the equation</p>
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 6\frac{\mathrm{d}y}{\mathrm{d}x} + 8y = x.\]
<h4 class="HELM_solution">Solution</h4>
<p>In the example before and the last question, we found that a fruitful approach for a first order ODE was to assume a solution in the same form as that on the right-hand side. Suppose we assume a solution \(y(x) = \alpha x\) and proceed to determine \(\alpha\). This appreach will actually fail, but let us see why.If \(y(x) = \alpha x\) then</p>
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha \qquad \text{and} \qquad \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0.\]
<p>Substitution into the differential equation yields</p>
\[0 - 6\alpha + 8\alpha x = \alpha.\]
<p>Comparing coefficients of \(x\):</p>
\[8\alpha x = x \qquad \text{so} \qquad \alpha = \frac{1}{8}\]
<p>Comparing constants:</p>
\[-6\alpha = 0 \qquad \text{so} \qquad \alpha = 0\]
<p>We have a contradiction! Clearly a particular integral of the form \(\alpha x\) is not possible. The problem arises because differentiation of the term \(\alpha x\) produces constant terms which are unbalanced on the right-hand side. So, we try a solution of the form \(y(x) = \alpha x + \beta\) with \(\alpha\), \(\beta\) constants. This is consistent with the recommendation in the table shown before. Proceeding as before</p>
\[\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha \qquad \text{and} \qquad \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0.\]
<p>Substitution in the differential equation now gives</p>
\[0 - 6\alpha + 8(\alpha x + \beta) = x.\]
<p>Equating coefficients of \(x\) and then equating constant terms we find:</p>
\[8\alpha = 1 \qquad \text{and} \qquad -6\alpha + 8\beta = 0\]
<p>From which we get \(\displaystyle \alpha = \frac{1}{8}\) and then \(\displaystyle \beta = \frac{3}{32}\).</p>
<p>The required particular integral is \(\displaystyle y_\text{p}(x) = \frac{1}{8}x + \frac{3}{32}\).</p>]]></text>
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