哪位大神帮忙解答一下

2024-12-23 14:16:11
推荐回答(1个)
回答1:

xdy-ydx=y²e^ydy
xy'-y=y²e^y
(xy'-y)/y²=e^y
凑微分,因为(x/y)'=(y-xy')/y²
所以
(x/y)'=-e^y
两边同时积分得
(x/y)=-e^y+C
x=-ye^y+Cy
x=y(C-e^y)