作者ashyan (今年的我沒有極限)
看板Grad-ProbAsk
標題Re: [理工] [工數]-高階ODE
時間Sat Aug 15 01:58:52 2009
※ 引述《winer8 (快來明星3 缺1 )》之銘言：
: y''+4y'+7y=13exp(-t)sin(t+1)
令 D = d/dx , t + 1 = u , -t = 1 - u
2 1-u
( D + 4D + 7 ) y = 13e sinu
2 1-u
yp = ( 1 / (D +4D+7) ) 13e sinu ( D = D-1 )
1-u 2 2
yp = 13e ( 1 / (D +2D+4) ) sinu ( D = -1 )
1-u
yp = 13e (1 / (2D+3) ) sinu
1-u 2
yp = 13e ( (2D+3) / (4D -9) ) sinu
1-u
yp = 13e ( (1/-13) (2cosu-3sinu))
1-u
yp = e ( 3inu - 2cosu )
-t
yp = e ( 3sin(t+1) - 2cos(t+1) )
: 答案是特解yp=[-2cos(t+1)+3sin(t+1)]exp(-t)
: (x^2D^2-2xD+2)y=x^3cosx
t 3t t
令 x = e => [ D(D-1) - 2D + 2 ] y = e cose
3t t
yp = ( 1 / (D-2)(D-1) ) e cose
t t -2t 3t t
yp = e S e S e e cose dt
t t t t
yp = e S e S e cose dt
t t t
yp = e S e sine dt
t t
yp = - e cose => yp = -xcosx
算完答案跟原PO的不一樣
我有翻了一下書
應該沒有算錯
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◆ From: 122.121.8.43
※ 編輯: ashyan 來自: 122.121.8.43 (08/15 02:24)
1F：推 winer8:SORRY第2題我是漏打負號了 請問第2題第2行怎變第第3行的呢 08/15 08:35
2F：推 youmehim:把 1/(D-2)(D-1) 拆成 1/(D-2) - 1/(D-1) 再代公式 08/15 09:53