[理工] [工數] 齊性 ODE
Consider the following initial value problem(IVP)
y'' + 2y' + y = 0 , y(0)=1 , y'(0)=c
where c is parameter. Find the range of c within which all solutions
of the given IVP are non-negative, that is,determine all possible
value of c which yield y(x)>=0 for x>=0.
ans: c>-1
我的作法
(D^2 + 2D + 1)y=0
(D+1)^2 y = 0
y=(c1+xc2)e^-x
y'=-c1e^-x+e^-xc2-xe^-xc2
y(0)=1=c1
y=(1+xc2)e^-x
y'(0)=c=-1+c2
=> c>= -1
為什麼不可以等於-1? c2=0不合嗎?
謝謝
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 114.37.53.41
※ 編輯: monkps 來自: 114.37.53.41 (04/07 17:36)