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# 請問這二題工數如何解?

1.y"'-y" +2y=0 2.y"'+8y=0

第2題解出來了，第1題還沒。

可是如何知道有 D = -1 的根?

sky-ccc很感謝您的幫忙,因為sam比較早回答,so ^"^ 分數也只能給一位,謝謝您~

以下是我準備要打的新問題：

主題是→ 解聯立方程式 若有興趣觀迎回答下一新的主題知識家。 ^^

3(x- 1 ) ^2 = 4(5- y) - (5- y) ^2 — ①

6(x - 1 ) = 4 + 2 ( y - 5) — ②

find x,y

### 3 Answers

- SamLv 610 years agoFavorite Answer
The characteristic polynomial is s^3-s^2+2=0, which has three roots, -1, 1+I and 1-i. The correspondent particular solutions is e^(-x), e^(1+i) and e^(1-i) in the complex form, ore^(-x), e^x*cos x and e^x*sin x in the real form.So the solution is Y= a*e^(-x)+b*e^x*cosx+c*e^x*sin x

2011-04-30 02:43:48 補充：

The factor theorem

The factor theorem states that a polynomial f(x) has a factor (x − k) if and only if f(k) = 0.

Let f(s)=s^3-s^2+2. it is easy to find that -1 is satisfying equation f(s)=0 by Observation.

By the factor theorem (s+1) is a factor of f(s).

2011-04-30 02:44:18 補充：

Use long division we can decompose f(s) as (s+1)(s^2-2s+2). Then use the formula of solution of the quadratic equation in radicals, we have the other two roots 1+i and 1-i.

- 10 years ago
y"'-y" +2y=0

可以找到

(m+1)(m^2 - 2m + 2)

m= -1 , 1+i ,1-i

三個根

y=C1 e^-x+ e^x(c2 cosx+c 3sinx)

希望有幫助到你~

2011-04-27 23:52:54 補充：

C3 我不小心打了空白健

y=C1 e^-x+ e^x(c2 cosx+c3 sinx)

2011-04-29 19:06:03 補充：

因式分解~~

把-1 帶入滿足這個可方程式

y"'-y" +2y=0

m^3-m^2+2=0

m代-1~~

Source(s): 自己