推 ntust661:推 03/17 04:38
※ 引述《chihhungw (涉世未深做人好難)》之銘言：
: (a) y"+9=r(t) with y(0)=y'(0)= 0
: r(t)=2u(t-1)-3δ(t-2)
Laplace轉換可得
9 2
s^2Y(s) + --- = ---e^-s - 3e^-2s
s s
2 3 9
=> Y(s) = -----e^-s - -----e^-2s - -----
s^3 s^2 s^3
-1 2 | | 9 2
=> y(t) = L {Y(s)} = t u(t)| - 3tu(t)| - ---t u(t)
|t->t-1 |t->t-2 2
2 9 2
= (t-1) u(t-1) - 3(t-2)u(t-2) - ---t u(t)
2
: (b) y"+2ty'-2y=0 ,y(0)=0 y'(0)=10 y(t)=?
Laplace轉換可得
d
[s^2Y(s) - sy(0) - y'(0)] - 2----[sY(s) - y(0)] -2Y(s) = 0
ds
d
=> (s^2Y(s) - 10 - 2Y(s) - 2s----Y(s) - 2Y(s) = 0
ds
d
=> (s^2-4)Y(s) - 2s----Y(s) = 10
ds
d s^2 - 4 -5
=> ----Y(s) + (---------)Y(s) = ---- 為一階O.D.E.
ds 2s s
求積分因子
2 s
I = exp[∫(--- - ---)ds]
s 2
1
= exp[2ln|s| - ---s^2]
4
-1
= s^2 e^----s^2
4
-5 -1 -1
IY(s) = ∫---- s^2 e^----s^2 ds = -5∫s e^---- ds (令s^2 = u , 2s ds = du)
s 4 4
1 -1 -1
= -5∫---e^----u du = 10e^----u
2 4 4
-1 1 1 10
=> Y(s) = 10e^----s^2 (-----e^---s^2) = -----
4 s^2 4 s^2
故所求
y(t) = 10t
因為說要詳細所以打很長= ="
: (c) Using the convolution theorem , solve
: y"+y =sint ,y(0)=0 y'(0)=0
Laplace轉換可得
1
(s^2 + 1)Y(s) = ---------
s^2 + 1
1 1
=> Y(s) = ---------*---------
s^2 + 1 s^2 + 1
∞
=> y(t) = sin t ＊ sin t = ∫ sin(τ)sin(t-τ)dτ
0
這邊偷偷用萊布尼茲作
1 1
L{---sin at} = --------- 兩端對a微分可得
a s^2 + 1
tcos at - sin at -2a
=> L{------------------} = ------------- 這邊a=1帶入
2a^2 (s^2 + a^2)
tcos t - sin t 1
=> L{----------------} = -----------
-2 (s^2 + 1)
故所求
-1
y(t) = sin t ＊ sin t = ----(tcos t - sin t)
2
: 感激不盡 希望能有詳細的步驟
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 112.105.128.71
※ 編輯: shinyhaung 來自: 112.105.128.71 (03/17 01:48)