Question

Differential-solve for y

(x-y)y'=x+y

Answer 1

so i started with the
\[Fx+Fy \frac{ dy }{ dx }=0\]
idea

Answer 2

In my class we use M=Fx and N=Fy
so,
M=(-x-y)
N=x-y

Answer 3

I keep getting \[My \neq Nx\]
-1=1

Answer 4

So i'm not sure where to go from here...i though maybe a V substitution..but it didn't seem to fit (maybe i missed it)

Answer 5

\[y'=\frac{x+y}{x-y}\]
Obviously, \(y=x\) is not a possible solution.

Substitute \(v=x-y\), so that \(v'=1-y'\):
\[y'=\frac{x+y}{x-y}~~\iff~~1-v'=\frac{2x-v}{v}\]
Rewritten, you have
\[v'=1-\frac{2x-v}{v}\]
Sorry, small mistake with the substituting