Differential Equation problems

I have tried really hard, but just couldn't figure these 2 problems out

1) (x-y)y'=x+y

2) (x+y)y'=1

1)
(x - y)y' = x + y

let y = ux; thus u = y/x
y' = u + xu'

(x - y)y' = x + y
(x - ux)(u + xu') = x + ux
(x - ux)(u + xu') = (x + ux)
x(1 - u)(u + xu') = x(1 + u)
(1 - u)(u + xu') = (1 + u)
(u + xu') = (1 + u) / (1 - u)
xu' = (1 + u) / (1 - u) - u
xu' = (1 + u²) / (1 - u)

separate variables ; just cross-multiply and change u' to du/dx
(1 - u) / (1 + u²) du = dx /x
[ 1/(1 + u²) - u /(1 + u²) ] du = dx/x
arctan(u) - ½ ln(1 + u²) = ln(x) + ln(c)
2arctan(u) - ln(1 + u²) = ln(Cx²)
2arctan(u) = ln[Cx²(1 + u²) ]
C(x² + y²) = e^[2arctan(y/x)]
------------------------------------- depending on how you simplify it.

2)
(x+y)y'=1
(x + y) dy/dx = 1
(x + y) = dx/dy

(x + y) = x'
x' - x = y

integrating factor = e^(∫-1 dy) = 1/e^y
x'/e^y - x/e^y = y/e^y

(x/e^y)' = y/e^y
(x/e^y) = ∫ y/e^y dy
(x/e^y) = -(1 + y) / e^y + constant
x = -1 - y + Ce^y
Ce^y - y = 1 + x

or

y - Ce^y = -(1 + x)
-------------------------also depends on how you simplify it

The first one is real simple.Divide both sides by (x-y).Then put y=kx for some k.
Put x+y=k in the second one and solve.I'm sure you'll get your answers!!