(x-y+1)y‘=1求微分方程通解.

(x-y+1)dy/dx=1得：dy/dx=1/(x-y+1)则：dx/dy=x-y+1,（1）将x看作函数,y看作自变量令z=x-y+1,则dz/dy=dx/dy-1因此（1）化为：dz/dy+1=z分离变量得：dz/(z-1)=dy两边积分得：ln(z-1)=y+lnC即：z-1=Ce^y将z=x-y+1代入...