I'm having trouble solving the following differential equation:

$y'(x)=\frac{8A^2x}{(1+4A^2x^2)^2}\cdot y-4Bx$

$A$ and $B$ are real constants.

I would be very grateful for any help. Thanks in advance.

**EDIT:**

A problem in classical mechanics led me to the Hamiltonian

$\mathfrak{H}=\frac{{p\_r}^2}{2m(1+4a^2r^2)}+\frac{{p\_{\phi}}^2}{2mr^2}+mgar^2$.

Therefore, I get $\dot{p\_r}=-\frac{\partial \mathfrak{H}}{\partial r}$ and
$\dot{p\_{\phi}}=-\frac{\partial \mathfrak{H}}{\partial \phi}$.

This yields $\dot{p\_{\phi}}=0\Rightarrow p\_{\phi}=$const.

and

$\dot{p\_r}=\frac{8ra^2{p\_r}^2}{2m{(1+4a^2r^2)}^2}+\frac{{p\_{\phi}}^2}{mr^3}-2mgar$.

If we assume $\dot{\phi}=0$, what we will from now on, then it follows that $p\_{\phi}=0$ and we get the simpler Hamiltonian
$\mathfrak{H}=\frac{{p\_r}^2}{2m{(1+4a^2r^2)}^2}+mgar^2$

and

$\dot{p\_r}=\frac{4ra^2{p\_r}^2}{m{(1+4a^2r^2)}^2}-2mgar$.

Since $\dot{p\_r}=m\dot{(\dot{r})}$, we get

$\dot{(\dot{r})}=\frac{4ra^2{\dot{r}}^2}{{(1+4a^2r^2)}^2}-2\cdot g\cdot a \cdot r$.

Let $z:=r$ and $A:=a$ and $B:=g\cdot a$.

Then we get

$z''=\frac{4zA^2{(z')}^2}{{(1+4A^2z^2)}^2}-2\cdot B \cdot z$.

We have $z''=\frac{dz'}{dt}=\frac{dz'}{dz} \cdot \frac{dz}{dt}=\frac{1}{2} \cdot \frac{d}{dz}({(z')}^2)$.

Set $p:=z'$. Then we get

$\frac{1}{2} \cdot \frac{d}{dz}({p}^2)=\frac{4zA^2{p}^2}{{(1+4A^2z^2)}^2}-2\cdot B \cdot z$

This finally led me to the problem

$y'(x)=\frac{8A^2x}{(1+4A^2x^2)^2}\cdot y-4Bx$.