differential equations

[Dell]

find y if

dy/dx=\frac{x-y+2}{x-y+3}

what i tried to do was
u=\frac{x-y+2}{x-y+3}


ux-uy+3u=x-y+2

y(1-u)=x+2-u(3+x)

y=\frac{x+2-u(3+x)}{(1-u)}


y'=\frac{(1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))}{1-2u+u^2}

\frac{(1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))}{1-2u+u^2}=u

(1-u'(3+x)-u)(1-u)+u'(x+2-u(3+x))=u-2u^2+u^3

from here try get u's one side anx x's the other
surely this isnt the way to do this ??

[Donaldos]

Try u=x-y.

[Dell]

perfect thanks

[Dell]

what would i do in a problem where the numerator and denominator have different x and y's


for example

y'=(3x-y-9)/(x+y+1)