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How should I start to solve the following differential equation? $$ xy''+(n-1)y'-Cxy^\frac{n+2}{n-2}=0, $$ where $x>0$, and $C$ is some constant.

I have very little knowledge in differential equations, tried some substitutions that did not work. I guess if the zero order term was not raised to the power, than it would be a more standard task, because we could re-write this to get a system of ODEs.

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Use the transformation $y(x) = \frac{w(x^{n-2})}{x^{n-2}}$, this will transform the ODE into $$ w''(\xi) = \frac{C}{(n-2)^2} \xi^{-2\frac{n-1}{n-2}} w(\xi)^{\frac{n+2}{n-2}}, $$ with $\xi = x^{n-2}$ (I might have made some mistakes, please check). This is the Emden-Fowler equation, and has particular solution $$ \left(\frac{(n-2)^2}{C}\right)^{\frac{n-2}{4}} \xi^{\frac{2-n}{2}}. $$ Hopefully, this will get you a bit further.

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Hint:

Case $1$: $n=1$ and $C\neq0$

Then $xy''-Cxy^{-3}=0$

$\dfrac{d^2y}{dx^2}=Cy^{-3}$

This reduces to an autonomous ODE.

Case $2$: $n\neq1,2$ and $C\neq0$

Then $xy''+(n-1)y'-Cxy^\frac{n+2}{n-2}=0$

$xy''-(1-n)y'=Cxy^\frac{n+2}{n-2}$

This belongs to a modified Emden–Fowler equation according to this.

Let $r=x^{2-n}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=(2-n)x^{1-n}\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left((2-n)x^{1-n}\dfrac{dy}{dr}\right)=(2-n)x^{1-n}\dfrac{d}{dx}\left(\dfrac{dy}{dr}\right)+(2-n)(1-n)x^{-n}\dfrac{dy}{dr}=(2-n)x^{1-n}\dfrac{d}{dr}\left(\dfrac{dy}{dr}\right)\dfrac{dr}{dx}+(2-n)(1-n)x^{-n}\dfrac{dy}{dr}=(2-n)x^{1-n}\dfrac{d^2y}{dr^2}(2-n)x^{1-n}+(2-n)(1-n)x^{-n}\dfrac{dy}{dr}=(2-n)^2x^{2-2n}\dfrac{d^2y}{dr^2}+(2-n)(1-n)x^{-n}\dfrac{dy}{dr}$

$\therefore(2-n)^2x^{3-2n}\dfrac{d^2y}{dr^2}+(2-n)(1-n)x^{1-n}\dfrac{dy}{dr}-(1-n)(2-n)x^{1-n}\dfrac{dy}{dr}=Cxy^\frac{n+2}{n-2}$

$(n-2)^2x^{3-2n}\dfrac{d^2y}{dr^2}=Cxy^\frac{n+2}{n-2}$

$\dfrac{d^2y}{dr^2}=\dfrac{Cx^{2n-2}y^\frac{n+2}{n-2}}{(n-2)^2}$

$\dfrac{d^2y}{dr^2}=\dfrac{Cr^{-\frac{2n-2}{n-2}}y^\frac{n+2}{n-2}}{(n-2)^2}$

Which reduces to an Emden–Fowler equation.

Let $\begin{cases}y=\dfrac{u}{s}\\r=\dfrac{1}{s}\end{cases}$ ,

Then $\dfrac{d^2u}{ds^2}=\dfrac{Cr^{-\frac{2}{n-2}-2}u^\frac{n+2}{n-2}}{(n-2)^2}$

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