Is there any way to find the this second order DE(contains y" and y^(-2))?

Question

While I was doing my Physics problems, I setup a equation contains $y''$ and $y^{-2}$. Specifically like this: $$my''=\frac{kQq}{(a-y)^2}-\frac{kQq}{(a+y)^2}$$ where $m,k,Q,q,a$ are constant. I did Laplace, but there's no formula for Laplace $y^{-2}$.

Is there any other way to solve this?

Answer 1

$$my''=\frac{kQq}{(a-x)^2}-\frac{kQq}{(a+x)^2}$$ $$my'=\frac{-kQq}{(a-x)}+\frac{kQq}{(a+x)}+ c $$ $$my={-kQq}\,{\,log (a-x)}+{kQq}\,{\,log(a+x)}+ cx + d $$ $$my={kQq} \quad log\frac{\,(a+x)}{\,(a+x)}+ cx + d $$

EDIT:

after question corrected..

$$ A= m/kQq$$

$$Ay''=\frac{4ay}{(a^2-y^2)^2}$$

Answer 2

$$C=\frac{kQq}{m}$$ $$y''=\frac{C}{(a-y)^2}-\frac{C}{(a+y)^2}=4aC\frac{y}{(y^2-a^2)^2}$$ $$2\:y'y''=4aC\frac{2y'y}{(y^2-a^2)^2}$$ $$y'=4aC\int \frac{2y}{(y^2-a^2)^2}dy=4aC\frac{1}{a^2-y^2}+c_1$$ $$dx=\frac{dy}{4aC\frac{1}{a^2-y^2}+c_1}$$ $$x=\int \frac{dy}{4aC\frac{1}{a^2-y^2}+c_1}=\int \frac{a^2-y^2}{4aC+c_1(a^2-y^2)}dy$$ $$x=\frac{y}{c_1}-4C\sqrt{\frac{a}{c_1^3(4C+c_1a)}}\tanh^{-1}\left(\sqrt{\frac{c_1}{a(4C+c_1a)}}\:y \right)+c_2$$ $x(y)$ is the solution expressed on the form of $x$ as a function of $y$.

The most likely, there is no closed form for the inverse function $y(x)$.