Solve the ordinary differential equation $\frac{d^2}{dx^2}F(x)=\frac{1}{F(x)^2}$ I've been trying to solve the ordinary differential equation
$$\frac{d^2}{dx^2}F(x)=\frac{1}{F(x)^2}$$
I tried simplifying and then simplifying even further and found that this function has to be in the form $e^{-cx}$ where $c$ is expressed in terms of the function itself therefore it means its a recurring function...So I need to know a function(that is not recurring) that is inversley proportional to its second derivative.I've  been trying to solve this for a long time but couldn't find anything online that was much help either.Help would be appreciated. Thanx in advance

As in Cesaro's comment, the first step is
$$2F'F''=\frac{2}{F^2} \implies (F')^2=c-\frac{2}{F}.$$ Then, separating variables, $$\int \sqrt{\frac{\phantom{^1}F\phantom{^1}}{cF-2}}\; dF = x+d$$ and the standard technique is to multiply by $F$ in numerator and denominator of the fraction inside the square root: $$x+d = \int \sqrt{\frac{F^2}{cF^2-2F}}\; dF =\int \sqrt{\frac{cF^2}{(cF-1)^2-1}}\; dF. $$ If $c$ is negative, say $c=-C$, it is clearer to write this as $$x+d = \int \sqrt{\frac{CF^2}{1-(CF+1)^2}}\; dF. $$ Now substitute $cF=1\pm \cosh \theta$ in the first case, and $CF=\sin \theta -1$ in the second case. If your application requires $F$ to be positive, then only the first case, with $cF=1 +\cosh \theta$, is needed. As Travis Willse says, you won't get a closed form for $F$ in terms of $x$, in general.
