Anonymous
Anonymous asked in 教育與參考考試 · 1 decade ago

method of reduction of order 解

1.xy''+2y'+xy=0,y1=x^(-1)cosx

2.(1-x^2)y''-2xy'+2y=0,y1=x

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  • 龍昊
    Lv 7
    1 decade ago
    Favorite Answer

    基本觀念:  以下公式是由 method of reduction of order 推導而來,建議上考場前記起來,以節省中間繁複計算的時間;若由題目可觀察到或給定一個解 y1,且方程式之型式如下:y'' + P(x)y' + Q(x)y = 0  則求解另一解 y2 之方法為:U = ( 1/y12 )e∫- P(x)dxy2 = y1∫Udx*  有了以上的觀念,我們就能來解算這個題目。*1. xy'' + 2y' + xy = 0 , given y1 = ( cos x )/x , find general solution of y = ?sol:  題目移項得:y'' + ( 2/x ) y' + y = 0  P(x) = ( 2/x )  Q(x) = 1  由題目給定 y1 = ( cos x )/x 為其中一解。  U = ( 1/y12 )e∫- P(x)dx   = ( x2/cos2x ) e∫- ( 2/x )dx   = ( x2/cos2x ) e - 2 ln│x│   = ( x2/cos2x ) ( 1/x2 )   = ( 1/cos2x )   = sec2x  y2 = y1∫Udx   = [ ( cos x )/x ]∫sec2x dx   = [ ( cos x )/x ] ( tan x )   = [ ( cos x )/x ] ( sin x/cos x )   = ( sin x )/x  通解 ( general solution ):y = c1y1 + c2y2  → y = c1[ ( cos x )/x ] + c2[ ( sin x )/x ] #*2. ( 1 - x2 )y'' - 2xy' + 2y = 0 , given y1 = x , find general solution of y = ?sol:  題目移項得:y'' + [ 2x/( 1 - x2 ) ] y' + [ 2/( 1 - x2 ) ] y = 0  P(x) = - 2x/( 1 - x2 )  Q(x) = 2/( 1 - x2 )  由題目給定 y1 = x 為其中一解。  U = ( 1/y12 )e∫- P(x)dx   = ( 1/x2 ) exp{∫[ 2x/( 1 - x2 ) ]dx }   = ( 1/x2 ) exp{ - ln│1 - x2│}   = ( 1/x2 ) [ 1/( 1 - x2 ) ]   = ( 1/x2 ) + ( 1/2 )[ 1/( 1 - x ) ] - ( 1/2 )[ 1/( 1 + x ) ]  y2 = y1∫Udx   = x∫{ ( 1/x2 ) + ( 1/2 )[ 1/( 1 - x ) ] - ( 1/2 )[ 1/( 1 + x ) ] }dx   = x [ ( - 1/x ) - ( 1/2 ) ln│1 - x│- ( 1/2 ) ln│1 + x│]   = x [ ( - 1/x ) - ( 1/2 ) ln│( 1 + x )( 1 - x )│]   = - 1 - ( x/2 ) ln│1 - x2│  通解( general solution ):y = c1y1 + c2y2  → y = c1x - k2[ 1 + ( x/2 ) ln│1 - x2│]  令 - k2 = c2  → y = c1x + c2[ 1 + ( x/2 ) ln│1 - x2│]  #*  以上過程應該都很詳細了;希望以上回答能幫助您。

    Source(s): 自己
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