How to solve $\frac{dy}{dx} = \frac{x^2-y^2}{x^2(y^2+1)}$ I tried to solve this using the solution of a first order differential equation but I don't think this can be reduced to that form.
How to approach this problem and find $y$?
Please help.
 A: Hint:
$\dfrac{dy}{dx}=\dfrac{x^2-y^2}{x^2(y^2+1)}$
$\dfrac{dx}{dy}=\dfrac{x^2(y^2+1)}{x^2-y^2}$
With reference to http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=180:
Let $u=\dfrac{x}{y}$ ,
Then $x=yu$
$\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u=\dfrac{y^2u^2(y^2+1)}{y^2u^2-y^2}$
$y\dfrac{du}{dy}=\dfrac{y^2(y^2+1)u^2}{y^2(u^2-1)}-u$
$y\dfrac{du}{dy}=\dfrac{y^2((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
$\dfrac{du}{dy}=\dfrac{y((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
Let $v=y^2$ ,
Then $\dfrac{du}{dy}=\dfrac{du}{dv}\dfrac{dv}{dy}=2y\dfrac{du}{dv}$
$\therefore2y\dfrac{du}{dv}=\dfrac{y((y^2+1)u^2-u(u^2-1))}{y^2(u^2-1)}$
$2\dfrac{du}{dv}=\dfrac{(v+1)u^2-u(u^2-1)}{v(u^2-1)}$
$(u^2v+u^2-u(u^2-1))\dfrac{dv}{du}=2(u^2-1)v$
$\left(v-u+1+\dfrac{1}{u}\right)\dfrac{dv}{du}=2\left(1-\dfrac{1}{u^2}\right)v$
This belongs to an Abel equation of the second kind.
Let $w=v-u+1+\dfrac{1}{u}$ ,
Then $v=w+u-1-\dfrac{1}{u}$
$\dfrac{dv}{du}=\dfrac{dw}{du}+1+\dfrac{1}{u^2}$
$\therefore w\left(\dfrac{dw}{du}+1+\dfrac{1}{u^2}\right)=2\left(1-\dfrac{1}{u^2}\right)\left(w+u-1-\dfrac{1}{u}\right)$
$w\dfrac{dw}{du}+\left(1+\dfrac{1}{u^2}\right)w=2\left(1-\dfrac{1}{u^2}\right)w+2u-2-\dfrac{4}{u}+\dfrac{2}{u^2}+\dfrac{2}{u^3}$
$w\dfrac{dw}{du}=\left(1-\dfrac{3}{u^2}\right)w+2u-2-\dfrac{4}{u}+\dfrac{2}{u^2}+\dfrac{2}{u^3}$
