Answer to Question #295402 in Differential Equations for Moni

Question

Solve by the method of variation of of parameter y''+y=secx

Answer 1

Solution:

Given $y^{\prime \prime}+y=\sec x$ .

The auxilary equation is $r^{2}+1=0$

Then, $r^{2}=\pm i \Rightarrow r=-i, r=i .$

Thus, the general solution is, $y_{c}=c_{1} \cos x+c_{2} \sin x .$

Here, $y_{1}(x)=\cos x$  and $y_{2}(x)=\sin x$ .

Then, $y_{1}^{\prime}(x)=-\sin x\ and\ y_{2}^{\prime}(x)=\cos x .$

Find the Wronskian of $y_{1}(x)=\cos x$  and $y_{2}(x)=\sin x .$

$\begin{aligned} W &=\left|\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right| \\ &=\cos ^{2} x+\sin ^{2} x \\ &=1 \end{aligned}$

Here, $f(x)=\sec x .$

$\begin{aligned} &u=-\int \frac{f(x) y_{2}(x)}{W(x)} d x \text { and } v=\int \frac{f(x) y_{1}(x)}{W(x)} d x \\ &u=-\int \frac{\sec x \sin x}{1} d x \quad \text { and } v=\int \frac{\sec x \cos x}{1} d x \end{aligned}$

$\begin{array}{ll} u=-\int \tan x d x & \text { and } v=\int 1 d x \\ u=\ln (\cos x) & \text { and } v=x \end{array}$

Thus, the particular solution is,

$y_{p}=\cos x \ln (\cos x)+x \sin x$

Therefore, the general solution of the given differential equation is,

$\begin{aligned} &y=y_{c}+y_{p} \\ &y=c_{1} \cos x+c_{2} \sin x+\cos x(\ln (\cos x))+x \sin x \end{aligned}$