**Unformatted text preview: **f ( x,y ) = C . From ∂f ∂x = 2 x + y 3 , we see that f ( x,y ) = x 2 + xy 3 + φ ( y ). From ∂f ∂y = 3 xy 2 and f ( x,y ) = x 2 + xy 3 , we see that 3 xy 2 = 3 xy 2 + φ ( y ), so φ ( y ) = 0. We can choose φ ( y ) to be 0 as well. (In general, φ ( y ) could be some constant K , but this will turn out not to matter.) The solution to the diﬀerential equation is given by f ( x,y ) = C , or x 2 + xy 3 = C. (Note that replacing x 2 + xy 3 with x 2 + xy 3 + K on the left-hand side doesn’t change anything, as both constants can be moved to the right-hand side.) The solution can further be rewritten as xy 3 = C-x 2 y 3 = C x-x y = ( C x-x ) 1 3 . 1...

View
Full Document

- Spring '07
- lee
- Derivative