Simple ordinary differential equation in r squared I need to solve this simple looking 1D problem, but I can't seem to express $r(t)$ explicitely
$$
\ddot{r}r^2-a=0
$$
where a is $a$ constant
(it originates from $F=G\frac{M_1M_2}{r^2}=M_1\ddot{r} $, the force between two pointmasses)
 A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
2\dot{r}\ddot{r} = {a \over r^{2}}\,2\dot{r}\quad\imp\quad
\totald{\dot{r}^{2}}{t} = -2a\,\totald{}{t}\pars{1 \over r}\quad\imp\quad
\totald{}{t}\pars{\dot{r}^{2} + {2a \over r}} = 0 
\end{align}
Then, $\dot{r}^{2} + 2a/r = 2a/r_{0}$ $\pars{~\mbox{assuming}\ a > 0~}$  and
$0 \leq r_{0} \leq r, \forall r$
\begin{align}
t + \overbrace{C}^{\mbox{constant}}
&=\int{\dd r \over \root{2a/r_{0} - 2a/r}}
=\root{r_{0} \over 2a}\int{\root{r}\,\dd r \over \root{r - r_{0}}} 
\\[3mm]&
=\root{r_{0} \over 2a}\
\overbrace{\int{\root{r_{0} + z^{2}}\pars{2z\,\dd z} \over z}}
^{r\ \equiv\ r_{0} + z^{2}}
=\root{2r_{0} \over a}\int\root{r_{0} + z^{2}}\,\dd z
\\[3mm]&=\root{2r_{0} \over a}\
\overbrace{\int\root{r_{0}}\sec\pars{\theta}\,\root{r_{0}}\sec^{2}\pars{\theta}
\,\dd\theta}
^{z\ \equiv\ \root{r_{0}}\tan\pars{\theta}}
\\[3mm]&=r_{0}\root{2r_{0} \over a}
\int\sec^{3}\pars{\theta}\,\dd\theta
\end{align}
Could you continue from here ?.
A: You can begin by dividing by $r^2$, then multiply by $\dot r$. You can now integrate the equation once to get a first order equation. In physical terms, this really boils down to the conservation of energy.
