None exact first order ODE i have to solve the following $1^{st}$ order differential equation
$(xy+1)dx+(2y-x)dy=0$
i am in the elementary differential class,and have not learned multivariate functions,
the equation below is none exact,since
$M_y=x\ne N_x=-1$ 
so i am looking for a substitution that can make it exact because the intergration factor has both $x$ and $y$ ...
$\large \frac{dy}{dx}=\frac{xy+1}{x-2y}$
 A: Approach $1$:
$(xy+1)~dx+(2y-x)~dy=0$
$(xy+1)~dx=(x-2y)~dy$
$(x-2y)\dfrac{dy}{dx}=xy+1$
Let $u=\dfrac{x}{2}-y$ ,
Then $y=\dfrac{x}{2}-u$
$\dfrac{dy}{dx}=\dfrac{1}{2}-\dfrac{du}{dx}$
$\therefore2u\left(\dfrac{1}{2}-\dfrac{du}{dx}\right)=x\left(\dfrac{x}{2}-u\right)+1$
$u-2u\dfrac{du}{dx}=\dfrac{x^2}{2}-xu+1$
$2u\dfrac{du}{dx}=(x+1)u-\dfrac{x^2}{2}-1$
$u\dfrac{du}{dx}=\dfrac{(x+1)u}{2}-\dfrac{x^2+2}{4}$
This belongs to an Abel equation of the second kind.
Let $s=x+1$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{ds}\dfrac{ds}{dx}=\dfrac{du}{ds}$
$\therefore u\dfrac{du}{ds}=\dfrac{su}{2}-\dfrac{(s-1)^2+2}{4}$
Let $t=\dfrac{s^2}{4}$ ,
Then $s=\pm2\sqrt t$
$\dfrac{du}{ds}=\dfrac{du}{dt}\dfrac{dt}{ds}=\dfrac{s}{2}\dfrac{du}{dt}$
$\therefore\dfrac{su}{2}\dfrac{du}{dt}=\dfrac{su}{2}-\dfrac{(s-1)^2+2}{4}$
$u\dfrac{du}{dt}=u-\dfrac{(s-1)^2+2}{2s}$
$u\dfrac{du}{dt}-u=\dfrac{(\pm2\sqrt t-1)^2+2}{\pm4\sqrt t}$
$u\dfrac{du}{dt}-u=\pm\sqrt t+1\pm\dfrac{3}{4\sqrt t}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
Approach $2$:
$(xy+1)~dx+(2y-x)~dy=0$
$(xy+1)~dx=(x-2y)~dy$
$(yx+1)\dfrac{dx}{dy}=x-2y$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-2y$
$yu\dfrac{du}{dy}+\dfrac{u}{y}=u-\dfrac{2y^2+1}{y}$
$yu\dfrac{du}{dy}=\dfrac{(y-1)u}{y}-\dfrac{2y^2+1}{y}$
$u\dfrac{du}{dy}=\dfrac{(y-1)u}{y^2}-\dfrac{2y^2+1}{y^2}$
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 $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$
$\therefore-\dfrac{1}{v^3}\dfrac{dv}{dy}=\dfrac{y-1}{y^2v}-\dfrac{2y^2+1}{y^2}$
$\dfrac{dv}{dy}=\dfrac{(2y^2+1)v^3}{y^2}-\dfrac{(y-1)v^2}{y^2}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
A: setting 
$y=vx$ we get 
$dy=vdx+xdv$ and substituting we get
$(vx^2+1)dx+(2y-1)(vdx+xdv)=0$
$(vx^2+2yv-vx+1)dx+(2yx-x^2)dv=0$
$(vx^2+2v^2x-vx+1)dx+x^2(2v-1)dv=0$
$(vx(x+2v-1)+1)dx+x^2(2v-1)dv=0$
this should be seperable but i cannot seperate it easily...
