展开全部
let
u = y/(x+1)^2
du/dx = [1/(x+1)^2]. dy/dx - 2y/(x+1)^3
dy/dx =(x+1)^2. { du/dx + [2/(x+1)] u }
/
y' - [2/(x+1)]y =(x+1)^4
(x+1)^2. { du/dx + [2/(x+1)] u } - 2(x+1)u =(x+1)^4
(x+1)^2. du/dx =(x+1)^4
du/dx =(x+1)^2
u = ∫ (x+1)^2 dx
=(1/3)(x+1)^3 + C
y/(x+1)^2 = (1/3)(x+1)^3 + C
y =(1/3)(x+1)^5 + C(x+1)^2
u = y/(x+1)^2
du/dx = [1/(x+1)^2]. dy/dx - 2y/(x+1)^3
dy/dx =(x+1)^2. { du/dx + [2/(x+1)] u }
/
y' - [2/(x+1)]y =(x+1)^4
(x+1)^2. { du/dx + [2/(x+1)] u } - 2(x+1)u =(x+1)^4
(x+1)^2. du/dx =(x+1)^4
du/dx =(x+1)^2
u = ∫ (x+1)^2 dx
=(1/3)(x+1)^3 + C
y/(x+1)^2 = (1/3)(x+1)^3 + C
y =(1/3)(x+1)^5 + C(x+1)^2
本回答被提问者采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询