Question

(y^2+2xy)dx-x^2dy=0
special integrating factors

Answer 1

i get that they aren't exact so use
\[My-Nx/N=2y+(2x/y)-2x/x^2\]
if i took my partial derivatives right

Answer 2

which then gives me
\[2y+(1/y)/x^2\]
which is one hell of an e^ to use for the integrating factor so im questioning my algebra

Answer 3

how 2x/y ?
shouldn't it be just 2x ?

Answer 4

\(\large \dfrac{\partial}{\partial y}2xy=2x\)

Answer 5

then your IF is just 2y/x^2

Answer 6

oh ok simple mistake. nice pic btw lol

Answer 7

ha, thanks ^_^