solve 2nd-order linear diff.!!

2y" + 3y' + y = 6e-x

Auxiliary equation:

2k2 + 3k + 1 = 0

(k + 1)(2k + 1) = 0

k = -1/2 or -1

So, yc = Ae-x/2 + Be-x

Set yp = Cxe-x

yp' = Ce-x - Cxe-x

yp" = -Ce-x - (Ce-x - Cxe-x) = Cxe-x - 2Ce-x

Plug in the equation, we get

2(Cxe-x - 2Ce-x) + 3(Ce-x - Cxe-x) + Cxe-x = 6e-x

-Ce-x = 6e-x

Comparing coefficients, C = -6

Therefore, y = yc + yp

= Ae-x/2 + Be-x - 6xe-x , where A and B are constants



2009-09-03 20:22:40 補充：
001只有答案，沒有計算過程，不算是真正的解答啊