Or can approach it this way:
[tex](2x^2+y)dx+(x^2yx)dy=0[/tex]
So, after a quick check for homogeneous, exact, and explicit calc. of an integrating factor via partials, we expand the differentials and attempt to group them together to form exact differentials:
[tex]2x^2dx+ydx+x^2ydyxdy=0[/tex]
Well, the ydxxdy can be grouped as:
[tex]y^2\left(\frac{ydxxdy}{y^2}\right)[/tex]
This leaves us with:
[tex]2x^2dx+x^2ydy+y^2 d\left(\frac{x}{y}\right)[/tex]
Can you rearrange this now to obtain exact differentials which can be integrated?
