Answer
$y=\int e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]dx+C_1$
Work Step by Step
We are given:
$y''+p(x)y'=q(x)$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$
Subsitute to the equation:
$\frac{du}{dx}+p(x)u=q(x)$
$\frac{u}{\sqrt 1+u^2}du=\frac{1}{a}dy$
Since this is a linear equation, we can get:
$u=e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]$
$\frac{dy}{dx}=e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]$
Integrating both sides:
$\int \frac{dy}{dx}=\int e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]$
$y=\int e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]dx+C_1$
The solution is:
$y=\int e^{-p(x)dx}[e^{-\int p(x)dx}q(x)dx+C]dx+C_1$
