Solving Differential Equation: Find y for dy/dx = (x-y+2)/(x-y+3)

[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 isn't 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'sfor example

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