Nonlinear Ordinary differential equation I would like to solve:
$$
\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$
with $y(0)=a$ and $y'(0)=0$
Where $a$ is a known constant. 
Thanks in advance. 
 A: Change variables so that the independent variable is $y$ and the dependent on $p=y'$. Then $p\dot p=y''$ (dots stand for derivatives w.r.t. $y$, primes w.r.t. $x$) and the equation becomes $$p\dot p-\frac2{y^2}=0$$ or $$\frac{d}{dy}(\frac{p^2}2)=\frac2{y^2}.$$ This can be solved by integrating into a first order equation for $y$ as a function of $x$.
A: Maybe this is the same technique, but what I have done before for $y''=f(y)$ is treated as $$\frac{{\rm d}y'}{{\rm d}x} = f(y) $$
$$\frac{{\rm d}y'}{{\rm d}y} \frac{{\rm d}y}{{\rm d}x} = f(y) $$
$$y'\,\frac{{\rm d}y'}{{\rm d}y} = f(y) $$
$$\int y'\,{\rm d}y' =\int f(y)\,{\rm d}y +K_0 $$
$$\frac{1}{2} y'^2 = g(y) $$ with $g(y)=\int f(y)\,{\rm d} y + K_0$
$$ y' = \sqrt{2\,g(y)} $$
$$ \frac{{\rm d}y}{\sqrt{2\,g(y)}} =  {\rm d}x$$
$$ x = \int \frac{1}{\sqrt{2\,g(y)}}\,{\rm d}y+K_1 $$
With your example 
$$f(y)=2/y^2$$ 
$$g(y)=\int 2/y^2\,{\rm d}y+K_0 = -2/y +K_0 $$
$$ x = \int \frac{1}{\sqrt{2\,\left(-2/y +K_0\right)}}\,{\rm d}y+K_1 $$
and with the IC given
$$ x = \frac{\sqrt{a}}{4} \left( 2 a \ln{\left(\sqrt{y-a}+\sqrt{y}\right)}-a \ln{a}+2 \sqrt{y (y-a)} \right) $$
A: Hint
$$\frac{d^2y}{dx^2} - \frac{2}{y^2}=0$$
Multiply by $2y'$
$$2y'\frac{d^2y}{dx^2} - \frac{4y'}{y^2}=0$$
$$((y')^2)' +4 \left(\frac{1}{y}\right)'=0$$
Then integrate
$$(y')^2 +\frac{4}{y}=K_1$$
$$.......$$
