Use the "mixed partials" check to see if the following differential equation is exact
(-1x^2 + 2y)dx + (2x - 3y^1)dy = 0
My(x,y) = ?
Nx(x,y) = ?
F(x,y) = ?
May not be what you want, but interesting to solve
$\displaystyle -x^2 dx+ (2ydx+2xdy) -3ydy= 0$
$\displaystyle \Rightarrow -x^2 dx+ 2d(xy) -3ydy= 0$
$\displaystyle \Rightarrow -\frac{x^3}{3}+2xy-\frac{3y^2}{2}=C$
Last edited by mr fantastic; Sep 18th 2009 at 08:01 AM.
Reason: Restored original post