推 Unis:原來如此 謝謝 06/26 16:53
※ 引述《Unis (阿芳)》之銘言:
: 求解如何反拉斯?!
: y"+y'+9y=0 y(0)=0.16 y'(0)=0
: => Y=0.16/(s^2+s+9)+0.16s/(s^2+s+9)
: 請問接下來該如何反拉思呢??
1/[s^2 + s + 9]
= 1/[(s + 1/2)^2 + ((√35)/2)^2]
=> exp(-t/2) sin((√35)t/2)
-------------------------
(√35)/2
s/[s^2 + s + 9]
= s/[(s + 1/2)^2 + ((√35)/2)^2]
= (s + 1/2)/[(s + 1/2)^2 + ((√35)/2)^2] - (1/2)/[(s + 1/2)^2 + ((√35)/2)^2]
exp(-t/2) cos((√35)t/2) exp(-t/2) sin((√35)t/2)
=> ----------------------- - (1/2) -------------------------
(√35)/2 (√35)/2
你應該一開始就將兩項合併
(s + 1)/[(s + 1/2)^2 + ((√35)/2)^2]
= (s + 1/2)/[(s + 1/2)^2 + ((√35)/2)^2] + (1/2)/[(s + 1/2)^2 + ((√35)/2)^2]
再做反拉普拉斯轉換
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