Re:[理工] [工數]-請問一題逆運算子
※ 引述《msu (do my best)》之銘言:
: 有一題
: (3) (1)
: y + y =2+2sinx
: ()表示階數
: 請問如何利用逆運算子求特解呢?
: 答案沒過程我看不太懂@@
: x
: yp= 2x - 2(---sinx)
: 2
: 感謝^^
yh不做了
yp我拆成兩個做
1 0 1
yp1 = ───── 2e = ── 2 = ∫ 2 dx = 2x
D(D^2+1) D
1 1 -2xcosx
yp2 = ──── 2sinx = ── ───── = ∫-xcosx = -xsinx -cosx
D(D^2+1) D 2
yp就是yp1+yp2 至於 -cosx 可以跟yh的 c1 cosx的未知項合併
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (10/12 23:48)
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (10/13 00:17)
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※ 編輯: CRAZYAWIND 來自: 59.105.159.190 (10/13 23:14)
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