find the general solution (x+y)y' = (3y-x), where y(1)=2

I did the substitution y=ux

I solved to get: ln|v-1| - 2/(v-1) = -ln|x| + A

I then rearranged and introduced u=y/x back into the equation:

ln|y-x| - 2x/(y-x) = A

For y i get

y = Ce^(2x/(y-x)) + x

I'm not too confident about this solution however, if anyone could check and see if im right i'd appreciate it, and if I'm wrong highlight where you think ive made an error :), thanks