dy/dx=(y-x)/(y+x)=(y/x-1)/(y/x+1),设y=xu,则dy/dx=u+xdu/dx,原方程化为u+du/dx=(u-1)/(u+1),整理得(u+1)du/(u^2+1)=-dx/x,两边积分∫u/(u^2+1)du+∫1/(u^2+1)du=-∫1/xdx1/2(ln(u^2+1))+arctanu=-lnx+lnC1u=y/x代...