Tedious Differential geodesic equation Suppose that you are given an autonomous second order non linear homogenious ODE thats the  only remaining equation of a system of ODEs, and it takes the form $$\frac{-2rr''}{r-r_s}=\frac{r_sr'^2}{(r-r_s)^2}+\frac{r_s\alpha_0}{r^2}$$, where $x'^n=\frac{d^nx}{dt^n}$. What would be the best way to solve this equation? Would this be solved as a normal autonomous differential equation?
 A: Letting $r(t)'=v(r)$, we see by the chain rule that
\begin{align}
r(t)''=\frac{\mathrm d^2r}{\mathrm dt^2}=\frac{\mathrm dr}{\mathrm dt}\frac{\mathrm dv}{\mathrm dr}=v\frac{\mathrm dv}{\mathrm dr}=v(r)v(r)'.
\end{align}
Notice that the independent variable has been changed to $r$ as apposed to $t$ now. Your ODE then becomes
\begin{align}
2vv'+\frac{r_s}{r(r-r_s)^2}v^2+r_s\alpha_0\frac{r-r_s}{r^3}=0.
\end{align}
Now note that the derivative of $v^2$ is $2vv'$, so we this equation is exact with the use of an integrating factor. We'll look for some factor $\mu(r)$ such that our equation can be written as
\begin{align}
(\mu(r) v^2)'+r_s\alpha_0\frac{r-r_s}{r^3}\mu(r)=0,
\end{align}
which can be integrated. To find $\mu(r)$ we'll expand the first part and equate like terms:
\begin{align}
2\mu(r)vv'+\mu(r)'v^2+...=2\mu(r)vv'+\frac{r_s}{r(r-r_s)^2}\mu(r)v^2+...
\end{align}
so then
\begin{align}
\mu(r)'=\frac{r_s}{r(r-r_s)^2}\mu(r).
\end{align}
I'll leave it to you to solve this ODE.
Now your equation can be written as
\begin{align}
\left(\mu(r)v^2\right)'+r_s\alpha_0\frac{r-r_s}{r^3}\mu(r)=0,
\end{align}
integrating and undoing the substitution for $r(t)'$ yields
\begin{align}
\mu(r)v^2+\alpha_0r_s\int\frac{r-r_s}{r^3}\mu(r)\mathrm dr=c_1,\\
(r(t)')^2=\frac{1}{\mu(r)}\left(c_1-\alpha_0r_s\int\frac{r-r_s}{r^3}\mu(r)\mathrm dr\right).
\end{align}
Then the solution to your equation is given by the integral
\begin{align}
\int\left[\frac{1}{\mu(r)}\left(c_1-\alpha_0r_s\int\frac{r-r_s}{r^3}\mu(r)\mathrm dr\right)\right]^{-1/2}\mathrm dr=t+c_2.
\end{align}
I'd be surprised if this has a known solution, let me know if you find it!
