Answer to Question #288871 in Differential Equations for Pankaj

Question

Using the method of variation of parameters, solve the equation:

d²y/dx² +a²y = sec ax.

Answer 1

$\text{ We have }\\ \left(D^{2}+a^{2}\right) y=\sec a x\\ \text{A.E. is } m^{2}+a^{2}=0 \Rightarrow m=\pm ai\\ \text{C.F. } =y_{c}=C_{1} \cos a x+C 2 \sin a x\\ y=A \cos a x+B \sin a x \ldots(1)\\ \text{ be the complete solution of the given equation where } A \text{ and } B \text{ are to be found.}\\ \text{We have, } y_{1}=\cos a x, y_{2}=\sin a x\\ y_{1}^{\prime}=-a \sin a x, y_{2}^{\prime}=a \cos a x\\ \begin{aligned} &W=y_{1} y_{2}^{\prime}-y_{2} y_{1}^{\prime}=a . \text { Also, } \phi(x)=\sec a x \\ &A^{\prime}=\frac{-y_{2} \phi(x)}{W}, \text { and } B^{\prime}=\frac{y_{1} \phi(x)}{W} \\ &A^{\prime}=\frac{-\sin a x \cdot \sec a x}{a}, \quad B^{\prime}=\frac{\cos a x \cdot \sec a x}{a} \end{aligned} \\[4mm] \begin{aligned} A^{\prime} &=\frac{-\tan a x}{a}, \quad B^{\prime}=\frac{1}{a} \\ B &=\frac{1}{a} \int d x+c_{2} \\ A &=-\frac{1}{a} \int \tan a x d x+C_{1}, \quad B=\frac{x}{a}+C_{2} \\ A &=\frac{-\log (\sec a x)}{a^{2}}+C_{1} \end{aligned} \\[4mm] \text{Substituting these values of } A \text{ and } B \text{ in Eqn. (1), we get: }\\$

$$\begin{aligned} y=& C_{1} \cos a x+C_{2} \sin a x &-\frac{\cos a x \log (\sec a x)}{a^{2}}+\frac{x \sin a x}{a} \end{aligned}$$