求微分方程(y^2-x^2)dy+2xydx=0的通解

dy/dx=2xy/(x^2-y^2)=(2y/x)/(1-(y/x)^2)
令y/x=u
y=ux,dy/dx=u+xdu/dx
所以
原式变为：
u+xdu/dx=2u/(1-u^2)
xdu/dx=(u+u^3)/(1-u^2)
(1-u^2)/(u+u^3)du=1/xdx
∫(1-u^2)/(u+u^3)du=∫1/xdx
解出即可.