I have a special solution for the Lane-Emden equation. Can I use it to find the general solution? The general Lane-Emden equation is $$\ddot{y}+\frac{2\dot{y}}{x}+y^N=0$$ where $y(0)=1$ and $\dot{y}(0)=0$.  If we eliminate the requirement that $y(0)=1$ there is a special solution for all real values of N, excluding N=1.
$$y=\bigg(\frac{xi}{\sqrt{\beta^2 +\beta}}\bigg)^\beta$$ 
where $\beta=\frac{2}{1-N}$.  
Is there some way to use this special solution to find the general solution?  It would be of great interest to astrophysicists.  
The equation is invariant to the Lie group $G(x,y)=(\lambda x,\lambda^\beta y)\lambda_o=0$.  When you apply this group to the DEQ you find that $\beta -2=N\beta$, or $\beta=\frac{2}{1-N}$  Its invariants (polynomial group stabilizers in the kernel of the map) are $\eta=\frac{\ddot{y}}{x^{\beta -2}}$, $\nu=\frac{\dot{y}}{x^{\beta-1}}$ and $\mu=\frac{y}{x^\beta}$.  The Lane-Emden equation expressed as invariants is $\eta+2\nu+\mu^N=0$.
$$
\frac{d\nu}{d\mu}=\frac{x\frac{d\nu}{dx}}{x\frac{d\mu}{dx}}=\frac{\eta-(\beta -1)\nu}{\nu-\beta\mu}
$$
Since $\eta=-\mu^N-2\nu$, this becomes
$$
(\mu^N+(1+\beta)\nu)d\mu+(\nu-\beta\mu)d\nu=0
$$ Solve this and you've got it.
The above special solution was found using the Lie algebra (also in the kernel of the map) between the invariants at singlarities, saddle points and along separatrices: $\eta=\beta(\beta -1)\mu$ and $\nu=\beta\mu$.  
 A: Hint:
Let $x=e^t$ ,
Then $t=\ln x$
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}=\dfrac{1}{x}\dfrac{dy}{dt}=e^{-t}\dfrac{dy}{dt}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(e^{-t}\dfrac{dy}{dt}\right)=\dfrac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\dfrac{dt}{dx}=\left(e^{-t}\dfrac{d^2y}{dt^2}-e^{-t}\dfrac{dy}{dt}\right)e^{-t}=e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}$
$\therefore e^{-2t}\dfrac{d^2y}{dt^2}-e^{-2t}\dfrac{dy}{dt}+2e^{-2t}\dfrac{dy}{dt}+y^N=0$
$e^{-2t}\dfrac{d^2y}{dt^2}+e^{-2t}\dfrac{dy}{dt}+y^N=0$
$\dfrac{d^2y}{dt^2}+\dfrac{dy}{dt}+e^{2t}y^N=0$
Let $y=e^{nt}u$ ,
Then $\dfrac{dy}{dt}=e^{nt}\dfrac{du}{dt}+ne^{nt}u$
$\dfrac{d^2y}{dt^2}=e^{nt}\dfrac{d^2u}{dt^2}+ne^{nt}\dfrac{du}{dt}+ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u=e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u$
$\therefore e^{nt}\dfrac{d^2u}{dt^2}+2ne^{nt}\dfrac{du}{dt}+n^2e^{nt}u+e^{nt}\dfrac{du}{dt}+ne^{nt}u+e^{2t}(e^{nt}u)^N=0$
$e^{nt}\dfrac{d^2u}{dt^2}+(2n+1)e^{nt}\dfrac{du}{dt}+n(n+1)e^{nt}u+e^{(Nn+2)t}u^N=0$
$\dfrac{d^2u}{dt^2}+(2n+1)\dfrac{du}{dt}+n(n+1)u+e^{((N-1)n+2)t}u^N=0$
Choose $n=\dfrac{2}{N-1}$ , the ODE becomes
$\dfrac{d^2u}{dt^2}+\dfrac{N+3}{N-1}\dfrac{du}{dt}+\dfrac{2(N+1)u}{(N-1)^2}+u^N=0$
Let $v=\dfrac{du}{dt}$ ,
Then $\dfrac{d^2u}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{du}\dfrac{du}{dt}=v\dfrac{dv}{du}$
$\therefore v\dfrac{dv}{du}+\dfrac{(N+3)v}{N-1}+\dfrac{2(N+1)u}{(N-1)^2}+u^N=0$
This belongs to an Abel equation of the second kind.
Let $v=-\dfrac{(N+3)w}{N-1}$ ,
Then $\dfrac{dv}{du}=-\dfrac{N+3}{N-1}\dfrac{dw}{du}$
$\therefore\dfrac{(N+3)^2w}{(N-1)^2}\dfrac{dw}{du}-\dfrac{(N+3)^2w}{(N-1)^2}+\dfrac{2(N+1)u}{(N-1)^2}+u^N=0$
$w\dfrac{dw}{du}-w=-\dfrac{2(N+1)u}{(N+3)^2}-\dfrac{(N-1)^2u^N}{(N+3)^2}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf or in http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf
