问一道求微分方程的题!我令t=x+1只能做到(3y-7t)dt+(7y-3t)dy=0(y-x+1)^2(y+x-1)^5=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 07:58:01
问一道求微分方程的题!我令t=x+1只能做到(3y-7t)dt+(7y-3t)dy=0(y-x+1)^2(y+x-1)^5=0
问一道求微分方程的题!
我令t=x+1
只能做到(3y-7t)dt+(7y-3t)dy=0
(y-x+1)^2(y+x-1)^5=0
问一道求微分方程的题!我令t=x+1只能做到(3y-7t)dt+(7y-3t)dy=0(y-x+1)^2(y+x-1)^5=0
令t=x-1 代入:(3y-7t)dt+(7y-3t)dy=0
dy/dt= -(3y-7t)/(7y-3t)= -(3y/t-7)/(7y/t-3)
令y/t=u y= ut, y'=u+tu'
t=x-1 代入:(3y-7t)dt+(7y-3t)dy=0
dy/dt= -(3y-7t)/(7y-3t)= -(3y/t-7)/(7y/t-3)
令y/t=u y= ut y'=u+tu'
u+tu'=-(3u-7)/(7u-3)
tu'=-(3u-7)/(7u-3)-u=(7-7u^2)/(7u-3)
(7u-3)du/(1-u^2)=7d...
全部展开
t=x-1 代入:(3y-7t)dt+(7y-3t)dy=0
dy/dt= -(3y-7t)/(7y-3t)= -(3y/t-7)/(7y/t-3)
令y/t=u y= ut y'=u+tu'
u+tu'=-(3u-7)/(7u-3)
tu'=-(3u-7)/(7u-3)-u=(7-7u^2)/(7u-3)
(7u-3)du/(1-u^2)=7dt/t
[2/(1-u)-5/(1+u)]du=7dt/t
-2ln(1-u)-5ln(1+u)=7lnt-lnC
(1-u)^2(1+u)^5t^7=C
(t-ut)^2(t+ut)^5=C
通
(x-1-y)^2(x-1+y)^5=C
收起
(3y-7x+7)dx+(7y-3x+3)dy=0
(3y-7(x-1))dx+(7y-3(x-1))dy=0
(3y-7(x-1))d(x-1)+(7y-3(x-1))dy=0
(3y-7u)du+(7y-3u)dy=0
(3y-7u)du+(7y-3u)dy=0
dy/du= -(3y-7u)/(7y-3u)
设y/u=p然后呢?我也是做到这一...
全部展开
(3y-7x+7)dx+(7y-3x+3)dy=0
(3y-7(x-1))dx+(7y-3(x-1))dy=0
(3y-7(x-1))d(x-1)+(7y-3(x-1))dy=0
(3y-7u)du+(7y-3u)dy=0
(3y-7u)du+(7y-3u)dy=0
dy/du= -(3y-7u)/(7y-3u)
设y/u=p
收起