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
 A: If we write $y = F(x)$ and regard $x$ as a function of $y$, the differential equation becomes
$$x'' = -\frac{(x')^3}{y^2},$$
where $'$ denotes $\frac{d}{dy}$. This equation is separable and first-order in $u := x'$: Rearranging gives
$$\frac{du}{u^3} = -\frac{dy}{y^2},$$
integrating gives
$$-\frac{1}{2 u^2} = \frac{1}{y} + C,$$
and rearranging again gives
$$(x')^2 = u^2 = - \frac{1}{\frac{1}{y} + C} .$$
We can solve for $x'$ and integrate the quantity in $y$ to give an explicit formulae for $x(y)$ in elementary functions. Note that the cases $C < 0$, $C = 0$, $C > 0$ yield qualitatively different solutions. For example, for $C > 0$,
$$\boxed{x = \frac{1}{2 \sqrt{2} C^{3 /2}}\left[2 \sqrt{-C y (C y + 1)} + \arcsin(2 C y + 1)\right] + D} .$$
Only in the $C = 0$ case can the solutions be inverted to give explicit formulas for $y = F(x)$ in terms of elementary functions. The real solutions in that case are
$$\boxed{F(x) = -\sqrt[3]{\frac{9}{2}}(\pm x - D)^{2/3}} .$$
Incidentally, you can find the last set of solutions quickly using the ansatz $F(x) = a (\pm x)^\lambda$, which immediately forces $\lambda = \frac{2}{3}$ and then determines $a$. Then, since the differential equation is autonomous, if $F(x)$ is a solution, so is $F(x - D)$ for any constant $D$.
A: It is a VERY difficult equation. If you are just interested for a solution try this by Wolfram:

A: 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.
A: $$\frac{d^2}{dx^2}F(x)=\frac{1}{F(x)^2}$$
Reduce the order of the DE:
$$2F'\frac{d^2}{dx^2}F(x)=2\frac{F'}{F(x)^2}$$
$$F'^2=-\frac{2}{F(x)}+C$$
it's now a first order separable DE.
