解初始值問題(非拉普拉斯變換方式)

This is very easy... Using the method of undetermining coefficients to tackle it.

y" + y = sin2t

Auxiliary equation: k2 + 1 = 0

k = +- i

Therefore, the complementary solution, yc = Asint + Bcost, where A and B are constants.

Now, let the particular solution,

yp = Csin2t + Dcos2t

yp' = 2(Ccos2t - Dsin2t)

yp" = -4(Csin2t + Dcost2t)

We have: -4(Csin2t + Dcos2t) + Csin2t + Dcos2t = sin2t

-3Csin2t - 3Dcos2t = sin2t

Comparing coefficients, C = -1/3, D = 0

So, the general solution:

y = yc + yp

= Asint + Bcost - 1/3 sin2t

y' = Acost - Bsint - 2/3 cos2t

y(0) = 2 and y'(0) = 1

So, B = 2, A - 2/3 = 1, A = 5/3

We have:

y = (5sint) / 3 + 2cost - (sin2t) / 3





2009-06-13 16:55:14 補充：
呢條喺Applied maths入面係執分題...

2009-06-13 17:07:23 補充：
這題用拉普拉斯變換方式做會很麻煩... 雖然我都做到同一個答案。
都係我答嘅方法最快...

我會比較鍾意用Variation of Parameters。

yes...我也試過用laplace去做...感覺上係繁複d...

2009-06-14 00:16:23 補充：
咁EMK你呢個方法係咪簡單很多??