对方程两边求微分,得(xdx+ydy)/(x^2+y^2)=[(xdy-ydx)/x^2]/(1+y^2/x^2),即 xdx+ydy=xdy-ydx,整理,得dy/dx=(x+y)/(x-y)。
两边对x求导得:(x+yy')/(x^2+y^2)=[(xy'-y)/x^2]/(1+y^2/x^2)x+yy'=xy'-yy'=(x+y)/(x-y)