ames asked in 教育與參考考試 · 1 decade ago

[數學] 請教幾題工程數學

1. Find the eigenvalues and eigenvectors of the matrix

-2 2 -3

B= 2 1 -6

-1 -2 0

2. Solve the initial value problem:

y\'=3x^2 - y/x , y(1)=5

3. Given f(t) = sin6tu s(t) , find the value of L[f\"(t)],(Hint us(t) is unit step fubction)

4. Given F(s)= 1 / s(s+2)^2 , find the value of L^-1[F(s)]

5. Given

A= 0 1

-2 -3

find the value of e^At

6. Find the solution of the O.D.E : y\' - y =1 + 3sint , y(0)=0

7. Find (t - 4t^2 y^3)dy + ( 4^t4 - y)dt = 0

1 Answer

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  • 1 decade ago
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    ●1.

    det(B-λI)=0 ,可得eigenvaluesλ= -3,-3,5

    λ= -3 代入 (B-λI)X=0 ,X=[x1,x2,x3]'

    可得eigenvectors [-2,1,0]',[3,0,1]'

    λ= 5 代入 (B-λI)X=0

    可得eigenvector [-1/2,1,1/2]'

    ●2.

    由I=∫e^(1/x)dx 可得一階方程式的積分因子I為 X

    代入原方程式 Iy=∫IQ,Q=3x^2

    y=3/4x^3 + c

    y(1)=5代入得 c = 17/4

    故得解 y=3/4x^3 + 17/4

    ●3.

    ●4.

    F(s)= 1 / s(s+2)^2 = 1/4s - 1/4(s+2) -1/2(s+2)^2

    L^-1[F(s)] = 1/4 -(e^-2t)/4 -(te^-2t)/2

    ●5.

    A eigenvalues 為 -2,-1 ,D=[-2 0;0 -1]

    得過渡矩陣S=[1,1;2,-1],S^-1=[1/3,1/3;2/3,-1/3]

    e^At = Se^DS^-1=1/3[(e^-2)+2(e^-1),(e^-2)-(e^-1);2(e^-2)-2(e^-1),2(e^-2)+(e^-1)]

    ●6.

    (1)

    由特性方程式可知m-1=0,m=1

    可得Yh=ce^x

    (2)

    再求Yp

    令Yp=A+Bsint+Ccost 代入原方程式

    可得A=1 ,B=C=-3/2

    y(x)=Yh+Yp=ce^x+1-(3/2)(sint+cost)

    再由y(0)=0,可得c=1/2

    故y(x)=(1/2)e^x+1-(3/2)(sint+cost)

    ●7. 這題題目有打錯嗎???

    (矩陣表示法我是用Matlab的語法喔,不知道你看不看的懂 如題目一的矩陣用Matlab來表示的話為[-2,2,-3;2,1,-6;-1,-2,0],[]' 如果矩陣右上方多一個 ' 的話表示轉置)

    2006-06-01 00:16:22 補充:

    ●3.L[f"(t)]=s^2 F(s)-sF(0)-F'(0)L[f"(t)]=6s/(s^2+36)

    Source(s): 不知道有沒有解錯^^"
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