113年台北科大電機碩士班丙組-工程數學詳解

國立臺北科技大學113學年度碩士班招生考試

系所組別: 電機工程系碩士班丙組
第一節 工程數學

解答: $$4\sin^2(y) \,dx-\sec^2(x)\,dy=0 \Rightarrow {1\over 4\sin^2(y)}\,dy={1\over \sec^2(x)}\,dx \\\Rightarrow -{1\over 4}\cot(y)= {1\over 2}x+{1\over 4}\sin(2x)+c_1 \Rightarrow \cot(y)=-2x-\sin(2x)+c_2\\ y({\pi\over 4})=1 \Rightarrow \cot(1)=-{\pi\over 2}-1+c_2 \Rightarrow c_2=\cot(1)+1+{\pi\over 2} \\ \Rightarrow \cot(y)=-2x-\sin(2x)+\cot(1)+1+{\pi\over 2} \\ \Rightarrow \bbox[red, 2pt]{y=\cot^{-1}\left(-2x-\sin(2x)+\cot(1)+1+{\pi\over 2} \right)}$$

解答: $$\textbf{(a)}\; y''-4y'+3y=0 \Rightarrow \lambda^2-4\lambda+3=(\lambda-3)(\lambda-1)=0 \Rightarrow \lambda=1,3 \\ \quad \Rightarrow \bbox[red, 2pt]{y_h =c_1e^x+c_2e^{3x}}\\ \textbf{(b)}\; \cases{y_1=e^x\\ y_2=e^{3x}} \Rightarrow W= \begin{vmatrix}y_1 & y_2 \\y_1' & y_2' \end{vmatrix} =2e^{4x} \\ \Rightarrow y_p =-e^x\int {e^{3x}(x+e^x+xe^x+e^{3x}) \over 2e^{4x}}\,dx +e^{3x} \int {e^{x}(x+e^x +xe^x+e^{3x}) \over 2e^{4x}}\,dx \\={4\over 9}+{1\over 3}x-{3\over 8}e^x -{1\over 4}e^{3x}-{3\over 4}xe^x-{1\over 4}x^2e^x +{1\over 2}xe^{3x} \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_3e^x+c_4e^{3x}+{4\over 9}+{1\over 3}x  -{3\over 4}xe^x-{1\over 4}x^2e^x +{1\over 2}xe^{3x}}$$

解答: $$f(t)=\begin{cases}0, & t\lt 2\\3, & 2\le t\lt 3\\ -3, &t\ge 3\end{cases} \Rightarrow f(t)=3(u(t-2)-u(t-3))-3u(t-3) =3u(t-2)-6u(t-3) \\ \Rightarrow L\{y''\}+ 6L\{y'\}+ 10L\{y\} =L\{3u(t-2)-6u(t-3)\} \\ \Rightarrow s^2Y(s)+6sY(s)+10Y(s)= {3e^{-2s} \over s}-{6e^{-3s}\over s} \Rightarrow Y(s)={3e^{-2s} \over s(s^2+6s+10)}-{6e^{-3s}\over s(s^2+6s+10)} \\ \Rightarrow y(t)=L^{-1}\{Y(s)\} =L^{-1}\left\{ {3e^{-2s} \over s(s^2+6s+10)}\right\}-L^{-1}\left\{ {6e^{-3s}\over s(s^2+6s+10)}\right\} \\\Rightarrow \bbox[red, 2pt]{y(t)=u(t-2) \left( {3\over 10}-{3\over 10}e^{-3(t-2)} (\cos(t-2)+3\sin(t-2) \right) }\\\qquad \qquad \bbox[red, 2pt]{-u(t-3) \left({3\over 5}-{3\over 5}e^{-3(t-3)}( \cos(t-3)+3\sin(t-3) \right)}$$

解答: $$A=\begin{bmatrix}-4 & 1 & 0 \\0 & -4 & -6 \\0 & 0 & -4\end{bmatrix} \Rightarrow sI-A=\begin{bmatrix}s& 0& 0 \\0 & s & 0\\ 0 & 0 & s \end{bmatrix}-\begin{bmatrix}-4 & 1 & 0 \\0 & -4 & -6 \\0 & 0 & -4\end{bmatrix}= \begin{bmatrix}s+4 & -1 & 0 \\0 & s+4 & 6 \\0 & 0 & s+4 \end{bmatrix} \\ \Rightarrow (sI-A)^{-1} = \begin{bmatrix}\frac{1}{s+4} & \frac{1}{(s+4)^2} & \frac{-6}{(s+4)^3} \\0 & \frac{1}{s+4} & \frac{-6}{(s+4)^2} \\0 & 0 & \frac{1}{s+4} \end{bmatrix} \Rightarrow L^{-1}\{ (sI-A)^{-1}\} = \bbox[red, 2pt]{\begin{bmatrix}e^{-4t} & te^{-4t} & -3t^2e^{-4t}\\0 & e^{-4t} & -6te^{-4t} \\ 0 & 0 & e^{-4t}\end{bmatrix}}$$

解答: $$\textbf{(a)}\; A={\sqrt 2\over 2} \begin{bmatrix}1 & -1 \\-1 & 1\\ 0 & 0 \end{bmatrix} =\begin{bmatrix}1/\sqrt 2 & -1/\sqrt 2 \\-1/\sqrt 2 & 1/\sqrt 2\\ 0 & 0 \end{bmatrix} \Rightarrow A^T= \begin{bmatrix}1/\sqrt 2 & -1/\sqrt 2 & 0\\-1/\sqrt 2 & 1/\sqrt 2 &0 \end{bmatrix} \\\Rightarrow B=A^TA =\begin{bmatrix}1 & -1 \\-1 & 1 \end{bmatrix} \Rightarrow \det(B-\lambda I)= \lambda(\lambda-2)=0 \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{0,2} \\ \textbf{(b)}\; C=AA^T= \begin{bmatrix}1 & -1 & 0 \\-1 & 1 & 0 \\0 & 0 & 0\end{bmatrix} \Rightarrow \det(C-\lambda I)=-\lambda^2(\lambda-2)=0 \Rightarrow \lambda=0,2\\ \lambda_1=0 \Rightarrow (C-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}1 & -1 & 0 \\-1 & 1 & 0 \\0 & 0 & 0\end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1=x_2 \\ \qquad \Rightarrow v= x_2 \begin{pmatrix}1   \\1\\ 0 \end{pmatrix} +x_3\begin{pmatrix} 0   \\0 \\ 1 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix} 1   \\1\\ 0 \end{pmatrix}, v_2= \begin{pmatrix} 0   \\0 \\ 1 \end{pmatrix} \\\lambda_2 =2 \Rightarrow (C-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}-1 & -1 & 0 \\-1 & -1 & 0 \\0 & 0 & -2\end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1 +x_2=0\\ x_3=0} \\\qquad \Rightarrow v= x_2 \begin{pmatrix} -1   \\1\\ 0 \end{pmatrix}, \text{choose }v_3= \begin{pmatrix} -1   \\1\\ 0 \end{pmatrix}\\ \Rightarrow \text{the expression of all eigenvectors of }AA^T = \bbox[red, 2pt]{  \left\{ s\begin{pmatrix}1   \\1\\ 0 \end{pmatrix} +t\begin{pmatrix} 0   \\0 \\ 1 \end{pmatrix}, k\begin{pmatrix} -1   \\1\\ 0 \end{pmatrix}, s,t,k \in \mathbb R\right\}}$$

 解答: $$\textbf{(a)}\;\text{Let }\cases{\vec v_1=(1,0,1)\\ \vec v_2=(2,1,0)\\ \vec v_3=(2,2,1)}, \text{ then by Gram-Schmidt process, we have}\\ \vec e_1={\vec v_1\over |\vec v_1|} =({1\over \sqrt 2}, 0, {1\over \sqrt 2})\\ \vec u_2=\vec v_2-(\vec v_2\cdot \vec e_1)\vec e_1=(2,1,0)-(1,0,1)=(1,1,-1) \Rightarrow \vec e_2={\vec u_2\over |\vec u_2|} =({1\over \sqrt 3},{1\over \sqrt 3},-{1\over \sqrt 3}) \\ \vec u_3= \vec v_3-(\vec v_3\cdot \vec e_1) \vec e_1-(\vec v_3\cdot \vec e_2)\vec e_2 =(-{1\over 2},1, {1\over 2}) \Rightarrow \vec e_3={\vec u_3\over |\vec u_3|} =(-{1\over \sqrt 6},{2\over \sqrt 6}, {1\over \sqrt 6}) \\ \Rightarrow \hat S=\{\vec e_1,\vec e_2,\vec e_3\} = \bbox[red, 2pt]{\{({1\over \sqrt 2}, 0, {1\over \sqrt 2}),({1\over \sqrt 3},{1\over \sqrt 3},-{1\over \sqrt 3}),(-{1\over \sqrt 6},{2\over \sqrt 6}, {1\over \sqrt 6}) \}} \\ \textbf{(b)} \; \mathbf x=a\vec e_1 +b\vec e_2 +c\vec e_3 \Rightarrow \begin{bmatrix} {1\over \sqrt 2} & {1\over \sqrt 3} & -{1\over \sqrt 6} \\0 & {1\over \sqrt 3} & {2\over \sqrt 6} \\{1\over \sqrt 2} & -{1\over \sqrt 3} & {1\over \sqrt 6} \end{bmatrix}  \begin{bmatrix} a\\b\\c \end{bmatrix} =\begin{bmatrix} 2\sqrt 2\\0\\\sqrt 2 \end{bmatrix} \equiv M\mathbf y=\mathbf x \\ \quad \Rightarrow \mathbf y=\begin{bmatrix}a\\b\\c \end{bmatrix} =M^{-1}\mathbf x = \begin{bmatrix}\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} & \frac{-\sqrt{3}}{3} \\\frac{-\sqrt{6}}{6} & \frac{\sqrt{6}}{3} & \frac{ \sqrt{6}}{6} \end{bmatrix} \begin{bmatrix}2\sqrt 2\\0\\\sqrt 2 \end{bmatrix} = \begin{bmatrix}3 \\\frac{\sqrt{6}}{3} \\ \frac{-\sqrt{3}}{3} \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\mathbf x= 3\vec e_1+ {\sqrt 6\over 3}\vec e_2-{\sqrt 3\over 3}\vec e_3}$$

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