Solución analítica para no Lineal de Segundo Orden ODE

Question

Estoy tratando de resolver el siguiente no lineal de segundo orden ODE donde $a$ e $b$ son constantes: $$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}-\frac{y}{ay+b}=0$$ 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:$$\frac{1}{x}\frac{d}{dx}\left(x\frac{dy}{dx}\right)=\frac{y}{ay+b}$$Cualquier sugerencia sería muy apreciada, gracias!

Answer 1

Sugerencia:

Suponga $a,b\neq0$ para el caso de la clave:

Deje $x=e^t$ ,

A continuación, $t=\ln x$

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

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

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

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

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