4 Finding a particular integral

In the previous subsection we explained what is meant by a particular integral. Now we look at a simple method to find a particular integral. In fact our method is rather crude. It involves trial and error and educated guesswork. We try solutions which are of the same general form as the $f\left(x\right)$ on the right-hand side.

Example 13

Find a particular integral of the equation

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-6y={\text{e}}^{2x}$

Solution

We shall attempt to find a solution of the inhomogeneous problem by trying a function of the same form as that on the right-hand side of the ODE. In particular, let us try $y\left(x\right)=A{\text{e}}^{2x},$ where $A$ is a constant that we shall now determine. If $y\left(x\right)=A{\text{e}}^{2x}$ then

$\frac{dy}{dx}=2A{\text{e}}^{2x}\text{and}\frac{d^{2}y}{d{x}^2}=4A{\text{e}}^{2x}.$

Substitution in the ODE gives:

$4A{\text{e}}^{2x}-2A{\text{e}}^{2x}-6A{\text{e}}^{2x}={\text{e}}^{2x}$

that is,

$-4A{\text{e}}^{2x}={\text{e}}^{2x}$

To ensure that $y$ is a solution, we require $-4A=1$ , that is, $A=-\frac{1}{4}$ .

Therefore the particular integral is $y_{\text{p}}\left(x\right)=-\frac{1}{4}{\text{e}}^{2x}$ .

In Example 13 we chose a trial solution $A{\text{e}}^{2x}$ of the same form as the ODE’s right-hand side. Table 2 provides a summary of the trial solutions which should be tried for various forms of the right-hand side.

Table 2 : Trial solutions to find the particular integral

Task!

By trying a solution of the form $y=\alpha {\text{e}}^{-x}$ find a particular integral of the equation   $\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-2y=3{\text{e}}^{-x}$

Substitute $y=\alpha {\text{e}}^{-x}$ into the given equation to find $\alpha$ , and hence find the particular integral:

Answer

Example 14

Obtain a particular integral of the equation:   $\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+8y=x.$

Solution

In Example 13 and the last Task, 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 $y\left(x\right)=\alpha x$ and proceed to determine $\alpha$ . This approach will actually fail, but let us see why. If $y\left(x\right)=\alpha x$ then $\frac{dy}{dx}=\alpha$ and $\frac{d^{2}y}{d{x}^2}=0$ . Substitution into the differential equation yields $0-6\alpha +8\alpha x=x$ and $\alpha$ .

Comparing coefficients of $x$ :

$8\alpha x=x\text{so}\alpha =\frac{1}{8}$

Comparing constants: $-6\alpha =0\text{so}\alpha =0$

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 $y\left(x\right)=\alpha x+\beta$ with $\alpha ,\beta$ constants. This is consistent with the recommendation in Table 2 on page 43. Proceeding as before $\frac{dy}{dx}=\alpha ,\frac{d^{2}y}{d{x}^2}=0.$ Substitution in the differential equation now gives:

$0-6\alpha +8\left(\alpha x+\beta \right)=x$

Equating coefficients of $x$ and then equating constant terms we find:

$8\alpha =1$ (1)

$-6\alpha +8\beta =0$ (2)

From $\left(1\right)$ , $\alpha =\frac{1}{8}$ and then from (2) $\beta =\frac{3}{32}$ .

The required particular integral is   $y_{\text{p}}\left(x\right)=\frac{1}{8}x+\frac{3}{32}.$

Task!

Find a particular integral for the equation:

$\frac{d^{2}y}{d{x}^2}-6\frac{dy}{dx}+8y=3cosx$

First decide on an appropriate form for the trial solution, referring to Table 2 (page 43) if necessary:

Answer

Answer

Equate coefficients of $cosx$ :

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Also, equate coefficients of $sinx$ :

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