※ 引述《a1133333 (阿傑)》之銘言：
: 求解
: y1'=3y1+y2-3sin3t
: y2'=7y1-3y2-16sin3t+9cos3t
Y' = AY + B , A = [ 3 1 ] B = [ - sin3t ]
[ 7 -3 ] [ -16sin3t + 9cos3t ]
from |A - λI| = 0
λ = 4, -4
when λ = 4, (A - λI)X1 = 0
[ -1 1 ][x1] = 0 , X1 = c1[ 1]
[ 7 -7 ][x2] [ 1]
λ = -4, (A - λI)X2 = 0
[ 7 1 ][x1] = 0 , X2 = c1[ 1]
[ 7 1 ][x2] [-7]
-1
Let S = [ 1 1 ] , S AS = D = [ 4 0 ]
[ 1 -7 ] [ 0 -4 ]
-1 -1
S = 1 [ 7 1 ] , S B = 1 [ -23sin3t + 9cos3t ]
---[ 1 -1 ] --- [ 13sin3t - 9cos3t ]
8 8
Let Y = SU , U = [ u1 ]
[ u2 ]
-1
Then Y' = AY + B => SU' = ASU + B => U' = DU + S B
u1' = 4u1 + (-23/8)sin3t + (9/8)cos3t
{
u2' = -4u2 + (13/8)sin3t - (9/8)cos3t
4t
u1 = c3e + (33/200)cos3t + (119/200)sin3t
{
-4t
u2 = c4e + (75/200)cos3t + (25/200)sin3t
Y = SU...
不知道有沒有錯 囧
遇到sin cos 我個人比較喜歡待定係數~
可是好像跟逆運算子差不多...
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※ 編輯: shinyhaung 來自: 112.104.36.205 (12/13 17:07)