Newton-Coulomb differential equation solution I am trying to solve the following DE:
$$
m \mathbf{\ddot{r}}=\frac{Q_1Q_2}{4\pi\epsilon_0\mathbf{r}\cdot\mathbf{r}}\mathbf{\hat{r}}
$$
Which I know is a tough one, so I settled for the simpler version:
$$
m \ddot{r}=\frac{Q_1Q_2}{4\pi\epsilon_0r^2}
$$
But I still couldn't figure out an exact solution to this DE, and can't really recall any method I know that would help but analytical solution which is not the goal, so I inserted it to Mathematica and got this equation after setting $c_1=1$ and $c_2=0$ for sake of simplicity:
$$
r \sqrt{1-\frac{\alpha }{r}}+\alpha \tanh ^{-1}\left(\sqrt{1-\frac{\alpha }{r}}\right)=t
$$
where:
$$
\alpha = \frac{Q_1Q_2}{2m\pi\epsilon_0}
$$
but the program fails at solving this equation and I too can't figure out a way to extract an exact solution.
So, if anyone can help, much appreciated! 
 A: That is an exact (albeit implicit) solution.  Sorry, you're not going to get a closed form expression for $r(t)$ as a function of $t$.
EDIT: The way to get the implicit solution is as follows.  This being an autonomous second-order differential equation, you can write a first-order equation for $v = dr/dt$ as a function of $r$:
$$ \dfrac{dv}{dr} = \dfrac{dv/dt}{dr/dt} = \dfrac{\alpha/(2r^2)}{v} = \dfrac{\alpha}{2 r^2 v}$$
This is a separable equation, so you get
$$\eqalign{2 v \; dv & = \alpha \;\dfrac{ dr}{r^2} \cr
       v^2 &= -\frac{\alpha}{r} + c_1\cr
       v &= \pm \sqrt{c_1 - \frac{\alpha}{r}}}$$
Writing $v = dr/dt$ again, that becomes another separable equation, and so
$$ t = \pm\int \dfrac{dr}{\sqrt{c_1 - \alpha/r}} + c_2 $$
and that rather complicated integral (in the case $c_1 = 1$) gives you Mathematica's solution.
Most integrals, and most differential equations, don't have closed-form solutions.  The question is not "why doesn't it?" but rather "why does it?" You're lucky to get even this much 
of a solution.
