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4x²ÿ 8xý-3y=7x²-15x³~請附詳解....3Q

2006-01-11 16:25:58 · 5 個解答 · 發問者 ? 3 in 科學 數學

5 個解答

1. Find the general solution of 4x2y'' + 8xy' - 3y = 7x2 - 15x3sol:  令 x = et → t = ln│x│  y' = dy/dx = ( dy/dt )( dt/dx ) = ( 1/x )( dy/dt )  y'' = d2y/dx2 = ( d/dx )[ ( 1/x )( dy/dt ) ]    = ( 1/x2 )( d2y/dt2 ) - ( 1/x2 )( dy/dt )  將 y'、y'' 代入原 Cauchy - Euler equation 得:  → 4( d2y/dt2 ) + 4( dy/dt ) - 3y = 7e2t - 15e3t  → ( d2y/dt2 ) + ( dy/dt ) - ( 3/4 )y = ( 7/4 )e2t - ( 15/4 )e3t  特徵方程式:λ2 + λ - 3/4 = 0  → ( λ - 1/2 )( λ + 3/2 ) = 0 → λ = 1/2 , - 3/2  yh = c1e( 1/2 )t + c2e( - 3/2 )t = c1x( 1/2 ) + c2x( - 3/2 ) ~ 齊性解  用未定係數法求特解:  Let yp = Ae2t + Be3t  → yp' = 2Ae2t + 3Be3t , yp'' = 4Ae2t + 9Be3t  ( d2yp/dt2 ) + ( dyp/dt ) - ( 3/4 )yp = ( 7/4 )e2t - ( 15/4 )e3t  → ( 21A/4 )e2t + ( 45B/4 )e3t = ( 7/4 )e2t - ( 15/4 )e3t  比較係數得:21A/4 = 7/4 → A = 1/3        45B/4 = - 15/4 → B = - 1/3  → yp = ( 1/3 )e2t - ( 1/3 )e3t = ( 1/3 )x2 - ( 1/3 )x3 ~ 特解  通解:y = yh + yp  → y = c1x( 1/2 ) + c2x( - 3/2 ) + ( 1/3 )x2 - ( 1/3 )x3 #

2006-01-12 07:41:00 補充:
實在搞不懂,沒給初值條件,甚至邊界值條件也未給,怎麼可以用Laplace轉換來算呢?太神奇啦!

2006-01-14 01:26:26 補充:
小頻版主,我就是要避免用參數變異法求特解,所以才這樣算,一方面積分減慢計算速度,令一方面要打行列式出來很麻煩,如果你要的話,請把e-mail給我,我寫一寫掃描圖檔給你比較快。

2006-01-12 02:39:19 · answer #1 · answered by 龍昊 7 · 0 0

那如果用郎斯基來求Yp怎麼求出解?

2006-01-13 19:27:00 · answer #2 · answered by ? 3 · 0 0

4x2y"+8xy'-3y=7x2-15x3(1)解齊次解 yh=?   4x2y"+8xy'-3y=0   齊次歐拉柯西方程式的解型式為 y=xm   代入可得輔助方程式: 4m(m-1)+8m-3=0    => 4m2+4m-3=0    => (2m-1)(2m+3)=0    => m=1/2, -3/2  故得 yh=C1x1/2+C2x-3/2(2) 求特解 yp=?    可使用未定係數法求解,    假設 yp=Ax3+Bx2+Cx+D         yp'=3Ax2+2Bx+C         yp"=6Ax+2B     代入原式     4x2(6Ax+2B)+8x(3Ax2+2Bx+C)-3(Ax3+Bx2+Cx+D)=7x2-15x3    => 45Ax3+21Bx2+5Cx-3D=7x2-15x3    比較係數可得 45A=-15, 21B=7, 5C=0, -3D=0    => A=-1/3, B=1/3, C=0, D=0    所以 yp=(-1/3)x3+(1/3)x2(3) 通解為     y=yh+yp=C1x1/2+C2x-3/2+(-1/3)x3+(1/3)x2

2006-01-12 07:07:40 · answer #3 · answered by 蔡春益 7 · 0 0

用未定係數法求特解我覺得也不是很好,太慢了!
直接用微分運算子才是快的方法!

2006-01-12 04:47:58 · answer #4 · answered by Tsai Chi-Nan 3 · 0 0

令t=lnx ; e^t= x
[4(D(D-1))+8D-3]y=7e^2t- 15e^3t
[4(D^2+D-3/4)]y=7e^2t- 15e^3t
(D^2+D-3/4)y=(7/4)e^2t- (15/4)e^3t
換成S領域 (S^2+S-3/4)Y(S)=N(S)+(7/4)/S-2 -(15/4)/S-3
Y(S)=N(S)/(S+3/2)(S-1/2)+[-(8/4)S+(9/4)]/ (S+2/3)(S-1/2)(S-2)(S-3)
Y(S)=C1/(S+3/2)+C2/(S-1/2)+A/(S-2)+B/(S-3)
A=1/3 ; B=-1/3
Y(S)=C1/(S+3/2)+C2/(S-1/2)+(1/3)/(S-2)+(-1/3)/(S-3)
Y(t)=C1e^-(3/2)t+C2e^(1/2)t+(1/3)e^2t-(1/3)e^3t
在轉成Y(x)=C1x^(-3/2)+C2x^(1/2)+1/3x^2-1/3x^3

2006-01-11 19:35:10 · answer #5 · answered by ? 2 · 0 0

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