How can solve this differential equation (second equation)? How can I solve this differential equation? 
$$
\frac{1}{h}\left(\frac{a}{\delta} - \frac{b}{\delta} - \frac{c}{E\delta^4}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left[A\,\frac{\mathrm{d}\delta}{\mathrm{d}x} \left(\frac{1}{\delta} + \frac{3B}{\delta^2}\right)\right],
$$
where $a,b,c,E,h,A,B$ are constants.
 A: Let $y=\dfrac{d\delta}{dx}$ ,
Then $\dfrac{1}{h}\left(\dfrac{a}{\delta}-\dfrac{b}{\delta}-\dfrac{c}{E\delta^4}\right)=\dfrac{d}{d\delta}\left(Ay\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)\right)\dfrac{d\delta}{dx}$
$\dfrac{1}{h}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)=A\dfrac{d}{d\delta}\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)y$
$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)d\delta$
$\int\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\int\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta^2}+\dfrac{3(a-b)B}{\delta^3}-\dfrac{c}{E\delta^5}-\dfrac{3Bc}{E\delta^6}\right)d\delta$
$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)^2\dfrac{y^2}{2}=-\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}+\dfrac{3(a-b)B}{2\delta^2}-\dfrac{c}{4E\delta^4}-\dfrac{3Bc}{5E\delta^5}\right)+c_1$
