Solving an inhomogenous parameter dependent ODE I was trying to solve the ODE
\begin{equation}
\ddot{r} r = \alpha(\dot{r}^2-1)
\end{equation}
where $\alpha$ is an arbitrary constant. There are some simple cases when $\alpha = -1 $ then you can use separation of variables to find the solution. For the initial condition when $r'(0)=1$ it also simplifies, $r(t)= t + r(0)$. Not sure how take on the general case. Any help is welcome!
 A: HINT:
\begin{equation}
\ddot{r} r = \alpha(\dot{r}^2-1)
\end{equation}
$$\frac{\dot{r}\ddot{r}}{\dot{r}^2-1}=\alpha\frac{\dot{r}}{r}$$
$$\frac{\dot{r}\ddot{r}}{(\dot{r}+1)(\dot{r}-1)}=\alpha\frac{\dot{r}}{r}$$
$$\frac{\ddot{r}}{\dot{r}+1} +\frac{\ddot{r}}{\dot{r}-1}=2\alpha\frac{\dot{r}}{r}$$
Then integrate both side 
A: Case $1$ : $\alpha=0$
Then $\ddot{r}r=0$
$r=0$ or $\ddot{r}=0$
$r=0$ or $r=C_1t+C_2$
$\therefore r=C_1t+C_2$
Case $2$ : $\alpha\neq0$
Then $\ddot{r}r=\alpha(\dot{r}^2-1)$
$r\dfrac{d^2r}{dt^2}=\alpha\left(\left(\dfrac{dr}{dt}\right)^2-1\right)$
Let $u=\dfrac{dr}{dt}$ ,
Then $\dfrac{d^2r}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=u\dfrac{du}{dr}$
$\therefore ru\dfrac{du}{dr}=\alpha(u^2-1)$
$\dfrac{u}{u^2-1}du=\dfrac{\alpha}{r}dr$
$\int\dfrac{u}{u^2-1}du=\int\dfrac{\alpha}{r}dr$
$\dfrac{1}{2}\ln(u^2-1)=\alpha\ln r+c_1$
$\ln\left(\left(\dfrac{dr}{dt}\right)^2-1\right)=2\alpha\ln r+c_2$
$\left(\dfrac{dr}{dt}\right)^2-1=C_1r^{2\alpha}$
$\dfrac{dr}{dt}=\pm\sqrt{C_1r^{2\alpha}+1}$
$dt=\pm\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$
$\int dt=\pm\int\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}$
$t=\pm\int_k^r\dfrac{dr}{\sqrt{C_1r^{2\alpha}+1}}+C_2$
