How to solve: $z''=-c/z^4$ I would like tho solve the following nonlinear second order differential equation
$\frac{d^2{z}}{d{t^{2}}}=-\frac{c}{z^4}$
Where $C$ is a constant.
I have try to use this steps http://www.sosmath.com/diffeq/second/nonlineareq/nonlineareq.html with no success.
Also, I've tryed to use the solver in Matlab, and they give me an error.
I don't think that this ODE is impossible to solve analytically.
Thank you in advance
 A: As rightly said by Claude Leibovici whom I salute, the solution would be "a true monster" if we want to express it on a closed form with elementary functions. To make it simpler, we need an inverse special function as shown below.
NOTE: The initial conditions are not specified in the wording of the question. Don't forget that with some initial conditions the arbitrary constants $C_1$ and $C_2$ might have particular values which make the result much simpler involving functions of lover level than the Beta and Inverse Beta functions.  

A: This is more a comment than an answer.
Using the classical $$\dfrac{d^2x}{dy^2}=-\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^3}\implies\dfrac{d^2y}{dx^2}=-\frac{\dfrac{d^2x}{dy^2}}{\left(\dfrac{dx}{dy}\right)^3 }$$ So, for the differential equation you posted $$\frac{d^2{z}}{d{t^{2}}}=-\frac{c}{z^4}\implies \frac{d^2{t}}{d{z^{2}}}=\frac{c}{z^4}\left(\frac{d{t}}{d{z}}\right)^3$$ Cheating (which means using a CAS), there is an explicit solution for $t$ as a function of $z$ but the result is $\color{red} {\text{a true monster}}$ which involves a bunch of elliptic integrals with very nasty arguments. I shall not try to type the solution here. May be, if you really want fun, try using Wolfran Alpha or Matlab.
