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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!

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    $\begingroup$ Maple finds no closed-form solution, and no symmetries. It is very likely that there are no closed form solutions (other than the trivial $y=0$). $\endgroup$ Commented Feb 7, 2019 at 21:55

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Hint:

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

Let $x=e^t$ ,

Then $t=\ln x$

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

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

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

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

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

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  • $\begingroup$ Is form this solvable? I'm still having trouble finding a solution method for this type of nonlinear equation. I found this additional transformation, but it doesn't seem to be working. $\endgroup$
    – Travis
    Commented Feb 11, 2019 at 21:01

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