Re: [工數] 為什麼重根時要多乘一個t

看板Math作者yueayase (scrya)時間10年前 (2014/04/23 02:40)推噓4(4推 0噓 8→)

留言12則, 6人參與討論串2/4 (看更多)

※ 引述《eolith123 (deer)》之銘言： : (4) 2 " 4 : y - 2k y + k y = 0 where k is constant : at : Let y = e : 4 2 2 4 : => a - 2k a + k = 0, a = ±k (k,k,-k,-k) : kt kt -kt -kt : y = Ae + Bte + Ce + Dte : 請問為什麼在重根時需要再乘上一個t? 可以用線性方程組的觀點來看 x1 = y [x1] ' , x =[x2] x2 = y [x3] '' [x4] x3 = y (3) x4 = y (4) 2 '' 4 (4) 2 4 因為 y - 2k y + k y = 0 <=> y = 2k y'' -k y [ 0 1 0 0 ] => x' = [ 0 0 1 0 ] x [ 0 0 0 1 ] 4 2 [-k 0 2k 0 ] 令A = [ 0 1 0 0 ] [ 0 0 1 0 ] [ 0 0 0 1 ] 4 2 [-k 0 2k 0 ] 對A對角化 4 2 2 4 2 2 2 2 2 det(A-cI) = 0 => c -2k c + k = (c - k ) = (c+k) (c-k) = 0 => c = -k, -k, k, k [ 1 ] [1 ] (A+kI)v = 0 => v = a[-k ] , (A-kI)v = 0 => v = b [k ] 2 2 [ k ] [k ] 3 3 [-k ] [k ] 所以兩個eigenvalue的eigenspace dimension都小於重根數 => 不可對角化但如果找A的Jordan form,則中間的那個J matrix應該是像 [k 1 0 0] J = [0 k 0 0] [0 0 -k 1] [0 0 0 -k] 所以原式變為 -1 -1 -1 x' = QJQ x <=> Q x' = JQ x -1 -1 ' 令 z = Q x => z' = Q x [z1] => z' = Jz , z = [z2] [z3] [z4] => z1' = kz1 + z2 z2' = kz2 z3' = -kz3 + z4 z4' = -kz4 kt kt kt -kt -kt -kt => z2 = c1e , z1 = c1 te + c2 e , z4 = c3 e , z3 = c3 te + c4 e 所以 kt kt [c1 te + c2 e ] x = Q kt [ c1 e ] -kt -kt [c3 te + c4 e ] -kt [ c3 e ] 所以 kt kt kt -kt -kt -kt y = x1 =q11(c1 te + c2 e ) + q12(c1 e ) + q13(c3 te + c4 e )+q14(c3 e ) kt kt -kt -kt = Ae + B te + C e + D te -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 220.129.27.62 ※ 文章網址: http://www.ptt.cc/bbs/Math/M.1398192056.A.8CD.html ※ 編輯: yueayase (220.129.27.62), 04/23/2014 02:42:09

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