113年中山資工碩士班-工程數學詳解

 國立中山大學113學年度碩士班招生考試

科目名稱: 工程數學【資訊工程礝士班乙組】

解答:$$A=\begin{bmatrix}1 & 0 & -2 \\0 & 5 & 0 \\ -2 & 0 & 4 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -\lambda^3+10\lambda^2-15\lambda =-\lambda(\lambda-5)^2 =0\\ \Rightarrow \text{ eigenvalues: }\bbox[red, 2pt]{0,5}$$

解答:$$B=\begin{bmatrix}-4 & 0 & 5 \\-3 & 3 & 5 \\-1 & 2 & 2 \end{bmatrix} \Rightarrow [B\mid I]= \left[ \begin{array}{rrr|rrr}-4 & 0 & 5 & 1 & 0 & 0 \\-3 & 3 & 5 & 0 & 1 & 0 \\-1 & 2 & 2 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1/(-4) \to R_1}\\\left[ \begin{array}{rrr|rrr}1 & 0 & \frac{-5}{4} & \frac{-1}{4} & 0 & 0 \\-3 & 3 & 5 & 0 & 1 & 0 \\-1 & 2 & 2 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_1+R_3\to R_3, 3R_1+R_2\to R_2} \left[ \begin{array}{rrr|rrr}1 & 0 & \frac{-5}{4} & \frac{-1}{4} & 0 & 0 \\0 & 3 & \frac{5}{4} & \frac{-3}{4} & 1 & 0 \\0 & 2 & \frac{3}{4} & \frac{-1}{4} & 0 & 1\end{array} \right] \\ \xrightarrow{R_2/3\to R_2} \left[ \begin{array}{rrr|rrr}1 & 0 & \frac{-5}{4} & \frac{-1}{4} & 0 & 0 \\0 & 1 & \frac{5}{12} & \frac{-1}{4} & \frac{1}{3} & 0 \\
0 & 2 & \frac{3}{4} & \frac{-1}{4} & 0 & 1 \end{array} \right] \xrightarrow{-2R_2+R_3\to R_3} \left[ \begin{array}{rrr|rrr} 1 & 0 & \frac{-5}{4} & \frac{-1}{4} & 0 & 0 \\0 & 1 & \frac{5}{12} & \frac{-1}{4} & \frac{1}{3} & 0 \\0 & 0 & \frac{-1}{12} & \frac{1}{4} & \frac{-2}{3} & 1 \end{array} \right] \\ \xrightarrow{-12R_3\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & \frac{-5}{4} & \frac{-1}{4} & 0 & 0 \\0 & 1 & \frac{5}{12} & \frac{-1}{4} & \frac{1}{3} & 0 \\0 & 0 & 1 & -3 & 8 & -12\end{array} \right] \xrightarrow{(-5/12)R_3+R_2 \to R_2, (5/4)R_3+R_1 \to R_1 } \\ \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -4 & 10 & -15 \\0 & 1 & 0 & 1 & -3 & 5 \\0 & 0 & 1 & -3 & 8 & -12 \end{array} \right] \Rightarrow B^{-1}= \bbox[red, 2pt]{\left[ \begin{array}{rrr|rrr}-4 & 10 & -15 \\1 & -3 & 5 \\-3 & 8 & -12\end{array} \right] }$$

解答:$$L\{x''\} +L\{x\} =L\{\sin(2t)\} \Rightarrow s^2X(s)-sx(0)-x'(0) +X(s)={2\over s^2+2^2} \\ \Rightarrow (s^2+1)X(s) ={2\over s^2+4}+2s+1 \Rightarrow X(s)={2\over (s^2+4)(s^2+1)} +{2s+1\over s^2+1} \\ =-{2\over 3}\cdot {1\over s^2+4}+{2\over 3} \cdot {1\over s^2+1}+{2s\over s^2+1} +{1\over s^2+1} =-{1\over 3}\cdot {2\over s^2+4}+{5\over 3}\cdot {1\over s^2+1} +2\cdot {s\over s^2+1} \\ \Rightarrow x(t) =L^{-1}\{X(s)\}\Rightarrow \bbox[red, 2pt]{x(t) =-{1\over 3}\sin(2t) +{5\over 3}\sin t+2\cos t}$$

解答:$$x'+2x=e^{-t} \Rightarrow \text{ integrating factor }I(t) =e^{2t} \Rightarrow e^{2t}x'+ 2xe^{2t}=e^t \Rightarrow (e^{2t}x)'=e^t \\ \Rightarrow e^{2t}x= \int e^t\,dt = e^t+c_1 \Rightarrow x=e^{-t}+c_1e^{-2t} \Rightarrow x(0)=1+c_1={3\over 4} \Rightarrow c_1=-{1\over 4} \\ \Rightarrow \bbox[red, 2pt]{x=e^{-t}-{1\over 4}e^{-2t}}$$

解答:$$\textbf{5.1}$$

$$\textbf{5.2}\;   a_k={1\over 2} \int_0^4 e^{-t} \cos({k\pi t\over 2})\,dt =\left. \left[ {e^{-t}k\pi \sin({k\pi t\over 2})-2e^{-t}\cos( {k\pi t\over 2}) \over k^2\pi^2 +4}\right] \right|_0^4 ={2(1-e^{-4})\over k^2 \pi^2+4} \\  \\\qquad \Rightarrow \bbox[red, 2pt]{a_k={2(1-e^{-4})\over k^2 \pi^2+4},k=0,1,2,\dots}$$

解答:$$\cases{y_1'=y_1+y_2+ 5\cos t\\ y_2'=3y_1-y_2-5\sin t} \Rightarrow \begin{bmatrix}y_1' \\y_2' \end{bmatrix} =\begin{bmatrix}1 & 1 \\3 & -1 \end{bmatrix} \begin{bmatrix}y_1 \\y_2 \end{bmatrix}+\begin{bmatrix} 5\cos t \\-5\sin t \end{bmatrix}\equiv \mathbf y'=A\mathbf y+\mathbf g \\ A= \begin{bmatrix}1 & 1 \\3 & -1 \end{bmatrix} \Rightarrow \text{eigenvalues of }A \text{ are 2,-2, and eigenvectors: } \begin{bmatrix}1\\ 1 \end{bmatrix}, \begin{bmatrix}1\\ -3 \end{bmatrix} \\ \Rightarrow \mathbb y_h=c_1e^{2t}\begin{bmatrix}1\\ 1 \end{bmatrix} +c_2e^{-2t} \begin{bmatrix}1\\ -3 \end{bmatrix} \Rightarrow \mathbf Y=\begin{bmatrix}e^{2t} & e^{-2t} \\e^{2t} & -3e^{-2t} \end{bmatrix} \Rightarrow \mathbf Y^{-1}= \begin{bmatrix}\frac{3}{4}e^{-2t} & \frac{1}{4} e^{-2t} \\\frac{1}{4}e^{2t} & -\frac{1}{4}e^{2t} \end{bmatrix} \\ \Rightarrow \mathbf Y^{-1}\mathbf g= \begin{bmatrix}\frac{3}{4}e^{-2t} & \frac{1}{4} e^{-2t} \\\frac{1}{4}e^{2t} & -\frac{1}{4}e^{2t} \end{bmatrix}  \begin{bmatrix} 5\cos t \\-5\sin t \end{bmatrix} =\begin{bmatrix} {15\over 4} e^{-2t} \cos t -{5\over 4}e^{-2t}\sin t \\ {5\over 4}e^{2t}\cos t+ {5\over 4}e^{2t} \sin t \end{bmatrix} \\ \Rightarrow \int \mathbf Y^{-1}\mathbf g\,dt =\begin{bmatrix} {5\over 4}e^{-2t}(-\cos t+\sin t) \\{1\over 4}e^{2t}(\cos t+ 3\sin t) \end{bmatrix} \\\Rightarrow \mathbf y_p=\mathbf Y \int \mathbf Y^{-1}\mathbf g\,dt = \begin{bmatrix}e^{2t} & e^{-2t} \\e^{2t} & -3e^{-2t} \end{bmatrix} \begin{bmatrix} {5\over 4}e^{-2t}(-\cos t+\sin t) \\{1\over 4}e^{2t}(\cos t+ 3\sin t) \end{bmatrix} =\begin{bmatrix}-\cos t+2\sin t \\-2\cos t-\sin t \end{bmatrix}\\ \Rightarrow \mathbf y=\mathbf y_h+\mathbf y_p =c_1e^{2t}\begin{bmatrix}1\\ 1 \end{bmatrix} +c_2e^{-2t} \begin{bmatrix} 1\\ -3 \end{bmatrix} +\begin{bmatrix}-\cos t+2\sin t \\-2\cos t-\sin t \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1(t) =c_1e^{2t}+ c_2e^{-2t}-\cos t+2\sin t\\ y_2(t)=c_1e^{2t} -3c_2e^{-2t}-2\cos t-\sin t}}$$

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