how to solve this differential equation $\frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$ by substitution? I've tried with this differential equation  $\displaystyle \frac{dy}{dx} = \frac{1+xy}{x(1-xy)}$  , put $u=xy$ then $\displaystyle\frac{du}{dx}=x\frac{dy}{dx}+y$ So,
It will be after editing 
$\displaystyle \frac{du}{dx} = \frac{1+u}{(1-u)} + \frac{u}{x}$
I am stuck here, keeping in mind only these three techniques are allowed: substitution, separating variable, and converting it to homogeneous equation
 A: Approach $1$:
$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$
$(x-x^2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{1}{x}-y$ ,
Then $y=\dfrac{1}{x}-u$
$\dfrac{dy}{dx}=-\dfrac{1}{x^2}-\dfrac{du}{dx}$
$\therefore x^2u\left(-\dfrac{1}{x^2}-\dfrac{du}{dx}\right)=x\left(\dfrac{1}{x}-u\right)+1$
$-u-x^2u\dfrac{du}{dx}=2-xu$
$x^2u\dfrac{du}{dx}=(x-1)u-2$
$u\dfrac{du}{dx}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$
This belongs to an Abel equation of the second kind.
Let $v=\dfrac{1}{x}$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{1}{x^2}\dfrac{du}{dv}$
$\therefore-\dfrac{u}{x^2}\dfrac{du}{dv}=\dfrac{(x-1)u}{x^2}-\dfrac{2}{x^2}$
$u\dfrac{du}{dv}=(1-x)u+2$
$u\dfrac{du}{dv}=\left(1-\dfrac{1}{v}\right)u+2$
Let $s=u-v$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}-1$
$\therefore(s+v)\left(\dfrac{ds}{dv}+1\right)=\left(1-\dfrac{1}{v}\right)(s+v)+2$
$(s+v)\left(\dfrac{ds}{dv}+\dfrac{1}{v}\right)=2$
$(s+v)\dfrac{ds}{dv}+\dfrac{s}{v}+1=2$
$(s+v)\dfrac{ds}{dv}=1-\dfrac{s}{v}$
$(v-s)\dfrac{dv}{ds}=v(s+v)$
Approach $2$:
$\dfrac{dy}{dx}=\dfrac{1+xy}{x(1-xy)}$
$(yx+1)\dfrac{dx}{dy}=x-yx^2$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-y\left(u-\dfrac{1}{y}\right)^2$
$yu\dfrac{du}{dy}+\dfrac{u}{y}=-yu^2+3u-\dfrac{2}{y}$
$yu\dfrac{du}{dy}=-yu^2+\dfrac{(3y-1)u}{y}-\dfrac{2}{y}$
$u\dfrac{du}{dy}=-u^2+\dfrac{(3y-1)u}{y^2}-\dfrac{2}{y^2}$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $u=\dfrac{1}{v}$ ,
Then $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=-\dfrac{1}{v^2}+\dfrac{3y-1}{y^2v}-\dfrac{2}{y^2}$
$\dfrac{dv}{dy}=\dfrac{v^3}{y^2}-\dfrac{(3y-1)v^2}{y^2}+v$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
