工程數學Second-order linear ODEs2？

計算過程和結果我都po在下面的網址
第一題:http://img23.imageshack.us/img23/4251/70772027.gif
第二題:http://img23.imageshack.us/img23/3391/80826827.gif
第三題:http://img23.imageshack.us/img23/843/31964573.gif

1. 10y" + 5y' + 0.625y = 0

輔助方程(Auxiliary equation): 10k2 + 5k + 0.625 = 0

解得k = -0.25 (重覆)

所以，y = Ae-0.25t + Bte-0.25t

y' = -0.25Ae-0.25t - 0.25Bte-0.25t + Be-0.25t

= (-0.25A + B - 0.25Bt)e-0.25t

y(0) = 2, y'(0) = -4.5

所以，A = 2, -0.25(2) + B = -4.5, B = -4

因此，y = (2 - 4t)e-0.25t



2. 20y" + 4y' + y = 0

輔助方程: 20k2 + 4k + 1 = 0

k = (-4 +- 8i) / 40 = -0.1 +- 0.2 i

所以，y = e-0.1t(Asin0.2t + Bcos0.2t)

y' = -0.1e-0.1t(Asin0.2t + Bcos0.2t) + 0.2e-0.1t(Acos0.2t - Bsin0.2t)

= 0.1e-0.1t[(2A - B)cos0.2t - (A + 2B)sin0.2t]

y(0) = 3.2, y'(0) = 1

所以，B = 3.2, 0.1(2A - 3.2) = 1, A = 6.6

因此，y = e-0.1t(3.2sin0.2t + 6.6cos0.2t)


3. y" = ky'

y" - ky' = 0

輔助方程: a2 - ka = 0

a = 0 或 k

因此，y = A + Bekt ，其中A和B是常數