Exact solution of 2nd order nonlinear ode

Question

I want to solve this ode $4a\frac{d^2y}{dx^2}=y^2+(b+2s)y-sx,$ where $a,b,s$ are fixed constants. There is no initial condition given and I want exact solution. Even any trivial solution will work. Please help me out.

Answer 1

Hint:

$4a\dfrac{d^2y}{dx^2}=y^2+(b+2s)y-sx$

$4a\dfrac{d^2y}{dx^2}=y^2+(b+2s)y+\dfrac{(b+2s)^2}{4}-sx-\dfrac{(b+2s)^2}{4}$

$4a\dfrac{d^2y}{dx^2}=\left(y+s+\dfrac{b}{2}\right)^2-sx-\dfrac{(b+2s)^2}{4}$

Let $u=y+s+\dfrac{b}{2}$ ,

Then $\dfrac{du}{dx}=\dfrac{dy}{dx}$

$\dfrac{d^2u}{dx^2}=\dfrac{d^2y}{dx^2}$

$\therefore4a\dfrac{d^2u}{dx^2}=u^2-sx-\dfrac{(b+2s)^2}{4}$

Let $t=x+\dfrac{(b+2s)^2}{4s}$ ,

Then $4a\dfrac{d^2u}{dt^2}=u^2-st$

Which relates to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=447.