(1+y)dx +(x-1)dy=0(1+y)dx =-(x-1)dy- ∫dx/(x-1) = ∫dy/(1+y)-ln|x-1| +C' =ln|1+y|(1+y)/(x-1) =e^C'1+y =C(x-1)y = C(x-1) -1
