Re: [理工] [工數][聯立ode]

看板Grad-ProbAsk作者ntust661 (Enstchuldigung~)時間15年前 (2011/02/28 13:43)推噓3(3推 0噓 1→)

留言4則, 4人參與討論串2/2 (看更多)

※ 引述《asdf322505 ()》之銘言： : y1'= -9y1+4y2-4y3 : y2'= -15y1+7y2-5y3 : y3'= 15y1-6y2+8y3 : ans; y1=2c1e^t+c3e^3t : y2=5c1e^t+c2e^2t : y3=c2e^2t-3c3e^3t : 這題怎嚜解? Y' = A Y mt Y = X e mt Y'= X m e mt mt X m e = A X e m X = A X (eigenvalue problem) -9-m 4 -4 │A - m│= [-15 7-m -5 ] 15 -6 8-m 3 3 2 = (-1) ( m - 6 m + ([-63+60]+[-72+60]+[56-30]) m - ([-9*7*8-20*15 -60*6 ]-[-60*7-60*8-30*9]) ) detA = (-504 - 300 - 360) + (420 + 480 + 270) = -1164 + 1170 = 6 3 3 2 故特徵方程式 (-1) (m - 6m + 11m - 6 ) = 0 m = 1 , m = 2 , m = 3 是根! 特徵向量呢 -10 4 -4 │A - 1│= [-15 6 -5 ] , Rank(A - I) = 2 15 -6 7 -15 x1 + 6 x2 - 7 x3 = 0 -15 x1 + 6 x2 - 5 x3 = 0 x1 = 6 , x2 = 15 , x3 = 0 6 2 特徵向量 X = [ 15 ] = [ 5 ] 0 0 -11 4 -4 │A - 2│= [-15 5 -5 ] , Rank(A - 2I) = 2 15 -6 6 -15 x1 + 5 x2 - 5 x3 = 0 -15 x1 + 6 x2 - 6 x3 = 0 x1 = 0 , x2 = 1 , x3 = 1 0 特徵向量 X = [ 1 ] 1 -12 4 -4 │A - 3│= [-15 4 -5 ] , Rank(A - 3I) = 2 15 -6 5 -15 x1 + 4 x2 - 5 x3 = 0 -15 x1 + 6 x2 - 5 x3 = 0 x2 = 0 , x1 = -1 , x3 = 3 -1 特徵向量 X = [ 0 ] 3 mt 故解 Y = X e 2 t 0 2t -1 3t Y = [ 5 ] c1 e + [ 1 ] c2 e + [ 0 ] c3 e 0 1 3 -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.45.251.207

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asdf322505

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passguestjay

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ntust661

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