Question

dy/dx=xy/x^2-y^2的通解怎么求,

Answer 1

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