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i have to solve the following $1^{st}$ order differential equation


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 $y$ ...

$\large \frac{dy}{dx}=\frac{xy+1}{x-2y}$

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Look into integrating factors. – Git Gud Oct 17 '13 at 23:08

Approach $1$:




Let $u=\dfrac{x}{2}-y$ ,

Then $y=\dfrac{x}{2}-u$






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 $s=\pm2\sqrt t$




$u\dfrac{du}{dt}-u=\dfrac{(\pm2\sqrt t-1)^2+2}{\pm4\sqrt t}$

$u\dfrac{du}{dt}-u=\pm\sqrt t+1\pm\dfrac{3}{4\sqrt t}$

This belongs to an Abel equation of the second kind in the canonical form.

Please follow the method in

Approach $2$:




Let $u=x+\dfrac{1}{y}$ ,

Then $x=u-\dfrac{1}{y}$


$\therefore yu\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-2y$




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 $\dfrac{du}{dy}=-\dfrac{1}{v^2}\dfrac{dv}{dy}$



Please follow the method in

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setting $y=vx$ we get

$dy=vdx+xdv$ and substituting we get





this should be seperable but i cannot seperate it easily...

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