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
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