Thread: Integrating factor View Single Post
 Sci Advisor HW Helper P: 1,593 Or can approach it this way: $$(2x^2+y)dx+(x^2y-x)dy=0$$ 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: $$2x^2dx+ydx+x^2ydy-xdy=0$$ Well, the ydx-xdy can be grouped as: $$y^2\left(\frac{ydx-xdy}{y^2}\right)$$ This leaves us with: $$2x^2dx+x^2ydy+y^2 d\left(\frac{x}{y}\right)$$ Can you re-arrange this now to obtain exact differentials which can be integrated?