※ 引述《jayemshow (S.Kazumi)》之銘言：
: 請問各位版友
: y"-4y'+ 4y = cos(t) ; y(0)=1 ; y'(0)=-1
: 這一題我怎麼算都跟答案不一樣 /_\
: 我最後是寫成這樣
: (As+B)[(s-2)^2]+C{[(S)^2]+1}+D{[(S)^2]+1}(s-2)=s
: 然後答案求出來是
: [(3/25)cos(t)]-[(4/25)sin(t)]+[(22/5)t(e^2t)]-[(3/25)(e^2t)]
: 不過正確答案是
: [(3/25)cos(t)]-[(4/25)sin(t)]-[(13/5)t(e^2t)]+[(22/25)(e^2t)]
: 不知道哪裡算錯了 @_@?
: 會不會是部份分式分錯了 XD
: 請各位版友提點一下
兩邊取Laplace transform
s
s^2 Y(s) - s + 1 - 4[s Y(s) -1] + 4 Y(s) = -----
s^2+1
s s^3-5s^2+2s-5
(s^2 -4s +4) Y(s) = ----- + s-5 = -------------
s^2+1 s^2+1
s^3-5s^2+2s-5
Y(s) = ---------------
(s^2+1)(s-2)^2
As+B C D
= ----- + ------- + -----
s^2+1 (s-2)^2 (s-2)
| -s+5+2s-5 s(3+4s) 3s-4 3s-4
As+B = (s^2+1)Y(s)| = --------- = ------------ = ---- = ----
| s^2=-1 -1-4s+4 (3-4s)(3+4s) 9+16 25
A=3/25, B=-4/25
| 8-20+4-5 -13
C = (s-2)^2 Y(s)| = -------- = ---
| s=2 4+1 5
d | (3s^2-10s+2)(s^2+1) - 2s(s^3-5s^2+2s-5) |
D = --- [(s-2)^2 Y(s)] | = ----------------------------------------|
ds | s=2 (s^2+1)^2 | s=2
(12-20+2)*5 - 4*(-13) 22
= --------------------- = ----
(4+1)^2 25
取inverse Laplace transform
y(t) = (3/25) cost - (4/25) sint - (13/5) te^(2t) + (22/25) e^(2t)
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