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.
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
Hints (I do not pretend to have a complete solution)
1) You can transform your equation in order to have a single parameter instead of 2:
Look for solutions of the form $y=\sqrt[4]{a}z$.
In this way, the given differential equation becomes:
$$\tag{1} \ z''+\dfrac{1}{z^3}=c \ \ \ \text{where} \ \ \ c:=\tfrac{b}{\sqrt[4]{a}} >0$$
2) A solution to the homogeneous equation associated with (1) (RHS = $0$) is:
$$z=\sqrt{2x}$$
