A second order nonlinear differential equation 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.
 A: 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.
A: 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 $\dfrac{d^2u}{ds^2}=\dfrac{Cr^{-\frac{2}{n-2}-2}u^\frac{n+2}{n-2}}{(n-2)^2}$
