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
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⤋