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

• You say the problem has solutions in terms of elementary functions : do you mean that you know solutions ? I am curious to know some of them, having spent some time on this issue yesterday. Commented Apr 3, 2017 at 1:11

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

• $(\frac{1}{2} (y')^2)' = y'y''$, so I see where you're getting that term, but why wouldn't integrating $y'y''$ with respect to $x$ result in $y'$? As $y'y'' = y'' \frac{dy}{dx}$. Multiplying by $dx$, you'd get $y''dy$, and integrating this would just become $y'$ Commented Apr 1, 2017 at 21:39
• The RHS should be $by+k$ Commented Apr 3, 2017 at 22:58

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:

$$z=\sqrt{2x}$$