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solve: x2y''-4xy'+4y=x4+x2過程越詳細越好^^Thanks!!

2006-10-15 12:16:00 · 2 個解答 · 發問者 土撥鼠 1 in 科學 數學

2 個解答

Problem:Solve  x2y'' - 4xy' + 4y = x4 + x2sol:  令 x = et → t = ln│x│  則:( dy/dx ) = ( dy/dt )( dt/dx )         = ( 1/x )( dy/dt )    ( d2y/dx2 ) = ( d/dx )( 1/x )( dy/dt )         = ( 1/x2 )( d2y/dt2 ) - ( 1/x2 )( dy/dt )  將以上代換結果代回原題目得到一新 o.d.e. 如下:  ( d2y/dt2 ) - 5( dy/dt ) + 4y = e4t + e2t  特徵方程式 ( characteristic equation ):r2 - 5r + 4 = 0  → ( r - 1 )( r - 4 ) = 0  → r = 1 , 4 ~ 相異實根  yh = c1et + c2e4t    = c1x + c2x4 ~ 齊次解 ( homogenous solution )  利用未定係數法 ( method of undetermined coefficient ) 求特解 yp。  令 yp = Ate4t + Be2t  → ( dyp/dt ) = Ae4t + 4Ate4t + 2Be2t    ( d2yp/dt2 ) = 8Ae4t + 16Ate4t + 4Be2t  ( d2yp/dt2 ) - 5( dyp/dt ) + 4yp = e4t + e2t  →  3Ae4t - 2Be2t = e4t + e2t  比較係數得:A = ( 1/3 )        B = - ( 1/2 )  → yp = ( te4t/3 ) - ( e2t/2 )      = ( x4 ln│x│/3 ) - ( x2/2 ) ~ 特解 ( particular solution )  通解 ( general solution ):y = yh + yp  → y = c1x + c2x4 + ( x4 ln│x│/3 ) - ( x2/2 ) #*  希望以上回答能幫助您。

2006-10-15 12:55:27 · answer #1 · answered by 龍昊 7 · 0 0

令x=e^t
t=lnx
d/dx=D
[D(D-1)-4D+4]y=e^4t+e^2t
令y=e^(mt)
m^2-5m+4=0
(m-4)(m-1)=0
yh=c1e^t+c2e^4t
將t改回來x
yh=c1x+c2x^4
[D^2-5D+4]y=e^4t+e^2t
我用heaveside逆運算子算
令yp=[1/(D-1)(D-4)]*(e^4t+e^2t)
yp=[1/(D-1)(D-4)]*(e^4t)+[1/(D-1)(D-4)]*(e^2t)
yp=e^4t*[1/D(D+3)]+e^t[1/(D)(D-3)]*e^t
yp=e^4t*[(1/3D)+(-1/9)+...]+e^t[1/(D-3)]*e^t
yp=e^4t*[(1/3D)+(-1/9)+...]+e^2t[1/(D-2)]
yp=e^4t*[(1/3D)+(-1/9)+...]+e^2t[(-1/2)+(-D/4)+...)]
∵上式的-1/9會跟齊次解衝到
則(-D/4)因為常數微分為零
所以不于理會
yp=(te^4t/3)+(-e^2t/2)
將t改回來x
yp=[(x^4lnx)/3]-(x^2/2)
y=yh+yp=c1x+c2x^4+[(x^4lnx)/3]-(x^2/2)
你如果想學這種方法
你再傳intemetkof99@yahoo.com.tw問我

2006-10-15 14:05:42 · answer #2 · answered by ? 3 · 0 0

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