(x-y+1)dy/dx=1得：dy/dx=1/(x-y+1)则：dx/dy=x-y+1 （1）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：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^yz=x-y+1代入得：x-y=Ce^y
