工數題目 求解

Method 1: Undetermined coefficients

y" + 10y' + 25y = e-5x

Auxiliary equation: k2 + 10k + 25 = 0, k = -5

So, the complementary solution: yc = Ae-5x + Bxe-5x

Let the particular solution, yp = Cx2e-5x

yp' = C(2xe-5x - 5x2e-5x) = C(2x - 5x2)e-5x

yp" = C[-5(2x - 5x2)e-5x + (2 - 10x)e-5x] = C(25x2 - 20x + 2)e-5x

So, C(25x2 - 20x + 2)e-5x + 10C(2x - 5x2)e-5x + 25Cx2e-5x = e-5x

2Ce-5x = e-5x

C = 1/2

So, the general solution: y = yc + yp = Ae-5x + Bxe-5x + 1/2 x2e-5x

y' = -5Ae-5x + Be-5x - 5Bxe-5x + xe-5x - 5/2 x2e-5x

y(0) = 1, A = 1

y'(0) = 0, -5 + B = 0, B = 5

So, y = (1 + 5x + 1/2 x2)e-5x




Method 2: By Laplace Transformation

y" + 10y' + 25y = e-5x

L{y" + 10y' + 25y} = L{e-5x}

[s2Y(s) - sy(0) - y'(0)] + 10[sY(s) - y(0)] + 25Y(s) = 1/(s + 5)

(s2 + 10s + 25)Y(s) - (s + 10) = 1/(s + 5)

(s2 + 10s + 25)Y(s) = (s2 + 15s + 51) / (s + 5)

Y(s) = (s2 + 15s + 51) / (s + 5)(s2 + 10s + 25)

Y(s) = 1/(s + 5) + 5/(s + 5)2 + 1/(s + 5)3

y = L-1{Y(s)}

= L-1{1/(s + 5) + 5/(s + 5)2 + 1/(s + 5)3}

= L-1{1/(s + 5) + 5(1!)/(s + 5)2 + 1/2 X (2!)/(s + 5)3}

= e-5x + 5xe-5x + 1/2 x2e-5x

= (1 + 5x + 1/2 x2)e-5x



2009-06-26 14:01:34 補充：
這只是partial fraction來的

Y(s) = (s^2 + 15s + 51) / (s + 5)(s^2 + 10s + 25) = (s^2 + 15s + 51) / (s + 5)^3

設它為A/(s + 5) + B/(s + 5)^2 + C/(s + 5)^3解出A, B與C值就可以了。

-.-.....
由特徵多項式 m^2 + 10 m + 25 = 0 可知齊次解為 (c1 + c2 x)*exp(-5x)

因外力向 exp(-5x) 包含於齊次式通解集合內，故應嘗試特解 yp = k (x^2) exp(-5x)，並代入題目可得下式：

　(k(x^2) exp(-5x))'' + 10 (k(x^2) exp(-5x))' + 25 (k(x^2) exp(-5x))
= 2k (exp(-5x)) = exp(-5x)　⇒　k = 0.5

2009-06-26 08:29:16 補充：
故通解為：y = (c1 + c2 x)*exp(-5x) + 0.5 (x^2) exp(-5x)
最後目視可知：由 y(0)=1　⇒　c1 = 1
　　　　　　　由y'(0)=0　⇒　c2 = 5

2009-06-26 08:30:51 補充：
PS：找特解利用 (uv)'' = u'' v+ 2u'v' + uv'' 心算。