Analytical Solution to Nonlinear Second Order ODE

Question

I'm trying to solve the following nonlinear second order ODE where $a$ and $b$ are constants: $$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}-\frac{y}{ay+b}=0$$ It looks somewhat like the modified Bessel equation, except the third term on the left makes it nonlinear. I've been trying to determine some way to find an analytical solution but haven't been able to come up with anything. It doesn't help much but it can also be written:$$\frac{1}{x}\frac{d}{dx}\left(x\frac{dy}{dx}\right)=\frac{y}{ay+b}$$Any suggestions would be greatly appreciated, thanks!

Answer 1

Hint:

Assume $a,b\neq0$ for the key case:

Let $x=e^t$ ,

Then $t=\ln x$

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{dy}{dt}=e^{-t}\dfrac{dy}{dt}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(e^{-t}\dfrac{d^2y}{dt^2}-e^{-t}\dfrac{dy}{dt}\right)e^{-t}=e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}$

$\therefore e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}+e^{-2t}\dfrac{dy}{dt}-\dfrac{y}{ay+b}=0$

$e^{-2t}\dfrac{d^2y}{dt^2}=\dfrac{y}{ay+b}$

$\dfrac{d^2y}{dt^2}=\dfrac{e^{2t}y}{ay+b}$