dy/dx=xy/(x^2-y^2) dx/dy=(x^2-y^2)/(xy)=x/y-y/x 设x/y=p x=py x'=p'y+p 代入原式得 p'y+p=p-1/p p'y=-1/p pdp=-dy/y 两边积分得 p^2=-lny+C 即 (x/y)^2=-lny+C
