How to solve $y'' + y = y^{-3}$

Question

I am trying to solve a central force problem and came to an equation like this:

$$\frac{d^2 y}{dx^2} + y = \frac{1}{y^3}$$

I can't find a decent method to solve it.

Answer 1

First of all, this equation has special solutions when $y'\equiv 0$ ($y=\pm 1, y=\pm i$). Let us look for such solutions that $y'$ does not vanish identically. Let us multiply both sides of the equation by $y'$:

$y'y''+yy'=\frac{y'}{y^3}$,

or

$\frac{1}{2}((y')^2)'+\frac{1}{2}(y^2)'=-\frac{1}{2}(\frac{1}{y^2})'$,

or

$(y')^2+y^2=-\frac{1}{y^2}+C$.

Now it is easy to separate $y$ and $x$.