How is the following differential equation solved? The equation is the following:
$$ {d^2 x(t) \over dt^2} = {k \over (x(t))^2} .$$
I used the software Maple to try and solve it, it can't. I did some research and this equation is obviously second order, and maybe of type "missing x", even if I don't know what that really means. I also found this substitution: $$ v = {d x(t) \over dt} $$ but nothing really came out of it. Also, none of my teachers are able to solve it, I have asked three of them. This equation comes the Newton's law of gravitation.
 A: This a second order autonomous ODE of the form $y'' = f(y)$. To solve it, see here. Basically, you
1) multiply by $x'(t)$ and integrate, to obtain $\frac{1}{2} x'^2 = -\frac{k}{x} + c$, where $c$ is a free constant. Then, you
2) rewrite this is a first order ODE, yielding $x' = \pm \sqrt{2 c - \frac{2 k}{x}}$. Now, you can
3) solve this ODE using separation of variables. You obtain the equation
\begin{equation}
\sqrt{2} c(t + t_0) = x(t)\sqrt{c - \frac{k}{x(t)}} + \frac{k}{2 \sqrt{c}} \log \left[2 c\, x(t)-k + 2 \sqrt{c}\,x(t) \sqrt{c - \frac{k}{x(t)}}\right],
\end{equation}
which you unfortunately cannot solve to obtain $x(t)$ as a function of $t$. 
A: We will rewrite the left hand side:
$$ \dfrac{d^2 x(t)}{dt^2} =\dfrac{d}{dt}\dfrac{d x(t)}{dt}=\dfrac{d}{dt}\dot{x}$$
Now use the chain rule (or symbolicly expand the fraction)
$$ =\dfrac{d\dot{x}}{dt}=\dfrac{d\dot{x}}{dx}\dfrac{dx}{dt}=\dfrac{d\dot{x}}{dx}\dot{x}$$
Pluging this into your ODE:
$$\dfrac{d\dot{x}}{dx}\dot{x}=\frac{k}{x^2(t)}$$
This is a separable ODE:
$$\dot{x}d\dot{x}=\frac{k}{x^2(t)}dx$$
Now you can solve this by simple integration:
$$\frac{1}{2}\dot{x}^2=-\frac{k}{x(t)}+\frac{c}{2}$$
Now solve for $\dot{x}$ and solve the separable ODE.

BTW Maple gave me a result:

with(DEtools):
ode := diff(x(t),t$2)-k/(x(t)^2)=0; 
dsolve(ode);

