1.10y''+5y'+0.625y=0,y(0)=2,y'(0)=-4.5
2.y''-9y=0,y(0)=-2,y'(0)=-12
3.20y''+4y'+y=0,y(0)=3.2,y'(0)=0
4.y''+2ky'+(k^2+w^2)y=0,y(0)=1,y'(0)=-k
5.y''-25y=0,y(0)=0,y'(0)=40
6.y''-2y'-24y=0,y(0)=0,y'(0)=20
工數二階常微分疑問,煩請工數高手為我解除疑惑。
2006-10-28 15:48:11 · 1 個解答 · 發問者 Anonymous in 教育與參考 ➔ 考試
1. 10y'' + 5y' + 0.625y = 0 , y(0) = 2 , y'(0) = - 4.5sol: 特徵方程式:10r2 + 5r + 0.625 = 0 → r = - 0.25 , - 0.25 ~ 重根 → y = ( c1 + c2x )e - 0.25x y(0) = 2 → x = 0 , y = 2 → c1 = 2 y' = c2e - 0.25x - 0.25( c1 + c2x )e - 0.25x y'(0) = - 4.5 → x = 0 , y' = - 4.5 → c2 - 0.25c1 = - 4.5 → c2 = - 4 → y = ( 2 - 4x )e - 0.25x #*2. y'' - 9y = 0 , y(0) = - 2 , y'(0) = - 1.2sol: 特徵方程式:r2 - 9 = 0 → r = ± 3 ~ 相異實根 → y = c1e - 3x + c2e3x y(0) = - 2 → x = 0 , y = - 2 → c1 + c2 = - 2 y' = - 3c1e - 3x + 3c2e3x y'(0) = - 1.2 → x = 0 , y' = - 1.2 → - 3c1 + 3c2 = - 1.2 解得:c1 = - 0.8、c2 = - 1.2 → y = - 0.8e - 3x - 1.2e3x #*3. 20y'' + 4y' + y = 0 , y(0) = 3.2 , y'(0) = 0sol: 特徵方程式:20r2 + 4r + 1 = 0 → r = - 0.2 ± 0.4 i ~ 共軛複根 → y = e - 0.2x ( c1cos 0.4x + c2sin 0.4x ) y(0) = 3.2 → x = 0 , y = 3.2 → c1 = 3.2 y' = - 0.2e - 0.2x ( c1cos 0.4x + c2sin 0.4x ) + e - 0.2x ( - 0.4c1sin 0.4x + 0.4c2cos 0.4x ) y'(0) = 0 → x = 0 , y' = 0 → - 0.2c1 + 04c2 = 0 → c2 = 1.6 → y = e - 0.2x ( 3.2 cos 0.4x + 1.6 sin 0.4x ) #*4. y'' + 2ky' + ( k2 + w2 ) y = 0 , y(0) = 1 , y'(0) = - ksol: 特徵方程式:r2 + 2kr + k2 + w2 = 0 → r = - k ± w i ~ 共軛複根 → y = ekx ( c1cos wx + c2sin wx ) y(0) = 1 → x = 0 , y = 1 → c1 = 1 y' = kekx ( c1cos wx + c2sin wx ) + ekx ( wc1sin wx + wc2cos wx ) y'(0) = - k → x = 0 , y' = - k → kc1 + wc2 = - k → c2 = - ( 2k/w ) → y = ekx [ cos wx - ( 2k/w ) sin wx ] #*5. y'' - 25y = 0 , y(0) = 0 , y'(0) = 40sol: 特徵方程式:r2 - 25 = 0 → r = ± 5 ~ 相異實根 → y = c1e - 5x + c2e5x y(0) = 0 → x = 0 , y = 0 → c1 + c2 = 0 y' = - 5c1e - 5x + 5c2e5x y'(0) = 40 → x = 0 , y' = 40 → - 5c1 + 5c2 = 40 解得:c1 = - 4、c2 = 4 → y = 4e5x - 4e - 5x #*6. y'' - 2y' - 24y = 0 , y(0) = 0 , y'(0) = 20sol: 特徵方程式:r2 - 2r - 24 = 0 → r = - 4 , 6 ~ 相異實根 → y = c1e - 4x + c2e6x y(0) = 0 → x = 0 , y = 0 → c1 + c2 = 0 y' = - 4c1e - 4x + 6c2e6x y'(0) = 20 → x = 0 , y' = 20 → - 4c1 + 6c2 = 20 解得:c1 = - 2、c2 = 2 → y = 2e6x - 2e - 4x #* 希望以上回答能幫助您。
2006-10-28 21:51:07 · answer #1 · answered by 龍昊 7 · 0⤊ 0⤋