求下列变量可分离方程的通解 (1)xy'-y=y^2(2)xy'-y^2+1=0

xy'-y=y^2
(xy'-y)/x^2=y^2/x^2
(y/x)'=(y/x)^2
两边积分得
-1/(y/x)=x+C
xy'-y^2+1=0
dy/(y^2-1)=dx/x
两边积分得
1/2ln(y-1)-1/2(y+1)=lnx+C1
即(y-1)/(y+1)=Cx^2