$y''+\frac{a}{y^3}=b$, with $a,b>0$. Solving a physics problem, this equation has arised:
$$y''-\frac{a}{y^3}=b$$
with $a,b>0$. Using a lagrangian I can avoid to solve this equation, but since the problem has a solution in terms of elementary functions, I want to know some way to solve this directly.
I did
$$y'y''-\frac{ay'}{y^{3}}=by' $$
and integrate, but eventually I get $y=0$.

Integrate both sides of the second form of your DE with respect to $x$ and then you're left with an equation
$$
\tfrac{1}{2} (y')^2 + a y^{-2}/2 = b y
$$
Then you just are left with a 1-st order DE from there
