If $xM(x,y)+yN(x,y)=0$, how to determinate the solution of the differential equation $M(x,y)dx+N(x,y)dy=0$?

Question

I've tried make the solution like this:

$xM(x,y)+xN(x,y)\frac{dy}{dx}=0$

and:

$yM(x,y)\frac{dx}{dy}+yN(x,y)=0$

Adding:

$xN(x,y)\frac{dy}{dx}+yM(x,y)\frac{dx}{dy}=0$

But i'm stucked here. I'm doing the right steps to get to the solution ?, thank you for your help.

Answer 1

If $xM(x,y)+yN(x,y)=0$, so $M(x,y)=-\dfrac{y}{x}N(x,y)$ and with equation $M(x,y)dx+N(x,y)dy=0$: $$-\dfrac{y}{x}N(x,y)dx+N(x,y)dy=0$$ you get $y=Cx$ from here.