$y''+\frac{a}{y^3}=b$, 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 $y=0$.

Substitute $y'=p(y)$ with the new unknown function $p$ of a variable $y$. Then separate variables.
