1. ## First Order Question

If i have:

(3y^2+2xy)dx-(2xy+x^2)dy=0

I firstly tried to solve this with Exact equations method. The partials I found were not equal therefore i assumed an integrating factor as a function of x and y. For both i did not come up with anything, as I got the factors as functions of both and x and y.

Then I proceeded to use the Homogenous Substituion v= y/x.
This lead me to getting a solution 2y+x=C.

I'm not sure if my approach is correct. I know for homogenous whenever the form y/x is recognizable, we proceed with this approach. But when is it correct to use the Exact Equations method?

Thanks.

2. you are in right direction .
$put\ y= vx \quad \Rightarrow \frac {dy}{dx}=v+x \frac {dv}{dx}$
then DE $(3y^2+2xy)dx=(2xy+x^2)dy \Rightarrow \frac{dy}{dx}=\frac{3y^2+2xy}{2xy+x^2}$
reduce to
$\frac {2v+1}{v(v+1)}dv= \frac{1}{x} dx$
resolve $\frac {2v+1}{v(v+1)}$ into partial fraction