TITLE: A second order differential equation, $y''=y^{-3}$
QUESTION [5 upvotes]: The exercise is about to solve $y''=1/y^3$ for $y=y(t)$ with initial values $y(0)=1$ and $y'(0)=2$.
My attempts:

I wrote up it as a first order diff.equation with $2$ variables by introducing $x:=y'$, so that we are looking for (the $y$-coordinate of) the trajectory of the vector field $(1/y^3, x)$, I could draw some vectors, and could 'guess' how the solution will look like.
Look for the solution in form $y=c\,t^\alpha$. I got that $\alpha=1/2$ and $c^4=-4$, that is $c=\pm1\pm i$. However, I guess, real function was asked.

I guess I'm missing something I should know about solving such a differential equation.

REPLY [3 votes]: Multiply $y'$,
$$
y'y''=y^{-3}y'$$
Integrate, 
$$
\frac 1 2 (y')^2 = -\frac 1 2 y^{-2} + c$$
Using initial condition, 
$$
2=-\frac 1 2 +c$$
Thus $c=\frac 5 2$. Thus, we have 
$$(y')^2+y^{-2}=5.$$
Multiply $y^2$, and substitute $Y=y^2$, then
$$\frac {(Y')^2} 4 +1 = 5Y,$$
$$(Y')^2=20Y-4$$
I think it is easy to go from here.