Ordinary differential equation

Characteristic equation: k^2 - 3k + 2 = 0
k = 1 or 2
Hence complementary solution: y = Ae^t + Be^2t
Let particular solution be y(p) = Cte^2t
Then y'(p) = 2Cte^2t + Ce^2t
y"(p) = 4Cte^2t + 4Ce^2t
So 4Cte^2t + 4Ce^2t - 6Cte^2t - 6e^2t + 2Cte^2t = 4e^2t
C = -2
So y = Ae^t + Be^2t - 2te^2t
y' = Ae^t + 2Be^2t - 4te^2t - 2e^2t
Put t = 0, -3 = A + B
5 = A + 2B - 2
Solving, A = -13 , B = 10.
So y = -13e^t + 10e^2t - 2te^2t

2010-11-08 07:21:19 補充：
這是微分方程,不能用普通方法做的。

d^2y/dt - 3dy/dt + 2y = 4e^(2t)
Integrate both sides wrt t :
dy/dt - 3y + 2ty = 2e^(2t) + C
5 - 3(-3) +2(0)(-3) = 2e^(2x0) (since y'(0)=5 y(0)=-3)
C = 12

Therefore,
dy/dt - 3y + 2ty = 2e^(2t) + 12
Integrate both sides wrt t again :
y - 3ty + yt^2 = e^(2t) + 12t + D
-3 - 3(0)(-3) + (-3)(0)^2 = e^(2x0) + 12(0) + D
D = -4

Therefore
y - 3ty + t^2y = e^(2t) +12t -4
y = [e^(2t) +12t -4]/(1-3t+t^2)