How do I know if a second order nonlinear differential equation is analytically solvable? How can I test if a non linear second order ODE is analytically solvable? 
for example:
$$y''+\frac{3}{x}y'=ay^3+by$$
where $a,b$ are constants, I've been trying to solve such an equation for a while but I'm starting to think that I should test if its solvable to begin with
 A: Hint:
Assume $a\neq0$ for the key case:
Let $x=e^t$ ,
Then $t=\ln x$
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{dy}{dt}=e^{-t}\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(e^{-t}\dfrac{d^2y}{dt^2}-e^{-t}\dfrac{dy}{dt}\right)e^{-t}=e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}$
$\therefore e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}+3e^{-2t}\dfrac{dy}{dt}=ay^3+by$
$e^{-2t}\dfrac{d^2y}{dt^2}+2e^{-2t}\dfrac{dy}{dt}=ay^3+by$
$\dfrac{d^2y}{dt^2}+2\dfrac{dy}{dt}=ae^{2t}y^3+be^{2t}y$
Let $y=e^{nt}u$ ,
Then $\dfrac{dy}{dt}=e^{nt}\dfrac{du}{dt}+ne^{nt}u$
$\dfrac{d^2y}{dt^2}=e^{nt}\dfrac{d^2u}{dt^2}+ne^{nt}\dfrac{du}{dt}+ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u=e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u$
$\therefore e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u+2e^{nt}\dfrac{du}{dt}+2ne^{nt}u=ae^{2t}e^{3nt}u^3+be^{2t}e^{nt}u$
$e^{nt}\dfrac{d^2u}{dt^2}+2(n+1)e^{nt}\dfrac{du}{dt}+n(n+2)e^{nt}u=ae^{(3n+2)t}u^3+be^{(n+2)t}u$
$\dfrac{d^2u}{dt^2}+2(n+1)\dfrac{du}{dt}+n(n+2)u=ae^{2(n+1)t}u^3+be^{2t}u$
