求微分方程Y'=Y/(Y-X)的通解dy/dx=(y/x)/[(y/x)-1].(1)令y/x=u,则y=ux,dy/dx=u+x(du/dx),代入(1)式得：u+x(du/dx)=u/(u-1),x(du/dx)=[u/(u-1)]-u=(2u-u²)/(u-1),分离变量得：[(u-1)/(2u-u²)]du=dx/x,积分...