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saltydog
#3
Sep25-05, 03:10 AM
Sci Advisor
HW Helper
P: 1,593
Or can approach it this way:

[tex](2x^2+y)dx+(x^2y-x)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^2ydy-xdy=0[/tex]

Well, the ydx-xdy can be grouped as:

[tex]y^2\left(\frac{ydx-xdy}{y^2}\right)[/tex]

This leaves us with:

[tex]2x^2dx+x^2ydy+y^2 d\left(\frac{x}{y}\right)[/tex]

Can you re-arrange this now to obtain exact differentials which can be integrated?