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/xdx
1/2(ln(u^2+1))+arctanu=-lnx+lnC1
u=y/x代入上式
ln根号下（y^2/x^2+1）+arctany/x=-lnx+lnC1
ln根号下(y^2+x^2)-ln根号下x^2+lnx-lnC1=-arctany/x
ln(C1*根号下(x^2+y^2))=-arctany/x
C1*根号下(x^2+y^2)=e^(-arctany/x)
根号下(x^2+y^2)=Ce^(-arctany/x)