Question

高等数学问题，解微分方程x^ydx=(1-y^2+x^2-x^2y^2)dy,麻烦写一下详细过程，多谢

Answer 1

题目应是 x^2*ydx=(1-y^2+x^2-x^2y^2)dy 吧。

x^2*ydx=(1-y^2+x^2-x^2y^2)dy 即 x^2*ydx=(1+x^2)(1-y^2)dy ,

则 [(1-y^2)/y]dy = [x^2/(1+x^2)]dx, 即 (1/y-y)dy = [1-1/(1+x^2)]dx,

lny-y^2/2 = x-arctanx +C.