y=(1+1n2x)(x+cosx),求 dy/dx

1个回答

Koutian飛

专业答主服务有保障

关注

展开全部

摘要您好，很高兴为您解答。y=(1+1n2x)(x+cosx)

dy/dx = 1/(x-y-1N2X)

dx/dy= x-y -1N2X

dx/dy -x = -y -1

The aux. equation

p-1=0

p= 1

let

xg= Ae^y

xp= By +C

xp' = B

xp'-xp=-y-1

B- (By +C) = -y-1

-By +(B-C) =-y-1

-B=-1 and B-C=-1

咨询记录 · 回答于2021-12-13

y=(1+1n2x)(x+cosx),求 dy/dx

您好，我这边正在为您查询，请稍等片刻，我这边马上回复您~

您好，很高兴为您解答。y=(1+1n2x)(x+cosx)dy/dx = 1/(x-y-1N2X)dx/dy= x-y -1N2Xdx/dy -x = -y -1The aux. equationp-1=0p= 1letxg= Ae^yxp= By +Cxp' = Bxp'-xp=-y-1B- (By +C) = -y-1-By +(B-C) =-y-1=>-B=-1 and B-C=-1

已赞过

评论收起