Flaming_Maniac wrote:
Yeah I know, we did integrating factors to make equations exact, but those integrating factors were by definition either functions of x or y. Not both.
well to be honest it was exactly the same. you just had to be more careful when you partially differentiated.
the reason you did u(xy) was that u(x,y) which is the "proper" way to do it for such a general DE but that's just way too fucking hard, so you do u(x), u(y) or something like that instead.
minty, did you try the problem? i got an answer which almost seems too straightforward for the second part of the question (the "hence solve (x
2 + xy + 1)dy + (y
2 + xy + 1)dx = 0" bit) - i got u(xy) = e
xy and the solution which i'll call G = e
xy(x+y) + c for any constant c. i mean, it
seems to work... but surely it's not that simple?
Last edited by Spark (2010-03-11 04:32:54)