111年中山資工碩士班-工程數學詳解

國立中山大學111學年度碩士班暨碩士在職專班招生考試

科目名稱:工程數學(資工系碩士班乙組)

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

解答:$$\textbf{2.1}\; A=\begin{bmatrix}-{\sqrt 3\over 2} & -{1\over 2} \\{1\over 2} & -{\sqrt 3\over 2} \end{bmatrix} =\begin{bmatrix}\cos{5\pi \over 6} & -\sin {5\pi \over 6} \\\sin {5\pi \over 6} & \cos {5\pi \over 6} \end{bmatrix} \\ \qquad \Rightarrow A\text{ is a rotation matrix which counterclock angle }\theta={5\pi \over 6}\\\textbf{2.2}\; {5\pi \over 6}\times {2011}= 837\times 2\pi +{11\pi \over 6} \Rightarrow A^{2011} =\begin{bmatrix}\cos{11\pi \over 6} & -\sin {11\pi \over 6} \\\sin {11\pi \over 6} & \cos {11\pi \over 6} \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}\sqrt 3/2 & 1/2 \\ -1/2 & \sqrt 3/2 \end{bmatrix}}$$

解答:$$\textbf{3.1}\; y'=ye^x-2e^x+y-2 =y(e^x+1)-2(e^x+1) =(y-2)(e^x+1) \\\quad \Rightarrow {1\over y-2}dy = (e^x+1)dx \Rightarrow \ln(y-2) =e^x+x+c_1 \Rightarrow y=e^{e^x+x+c_1}+2   \Rightarrow \bbox[red, 2pt]{y=c_2e^{e^x+x}+2} \\ \textbf{3.2}\; ty'=t^4-2y \Rightarrow y'+ {2\over t}y=t^3 \Rightarrow \text{ integrating factor }I(x)=e^{\int (2/t)\,dt} = t^2 \\\quad \Rightarrow   t^2 y'+ 2t y=t^5 \Rightarrow (  t^2y)'=t^5 \Rightarrow t^2y={1\over 6}t^6+c_1 \Rightarrow y={1\over 6}t^4+{c_1\over t^2} \\ y(1)=0 \Rightarrow 0={1\over 6}+c_1 \Rightarrow c_1=-{1\over 6} \Rightarrow \bbox[red, 2pt]{ y={1\over 6}t^4-{1\over 6t^2}}$$

解答:$${a_0\over 2}={1 \over 2\pi} \int_{-\pi}^\pi f(x)\,dx \Rightarrow a_0={1\over \pi} \left(\int_{-\pi}^0 x\,dx +\int_0^\pi -x\,dx  \right)={\pi^2\over \pi} =\pi \\ a_n= {1\over \pi} \int_{-\pi}^\pi f(x) \cos(nx)\,dx ={1\over \pi} \left(\int_{-\pi}^0 x \cos(nx) \,dx +\int_0^\pi -x \cos(nx)\,dx  \right) \\\qquad ={1\over \pi} \cdot {2\over n^2}(1-(-1)^n) \Rightarrow a_1={4\over \pi} \\\Rightarrow \bbox[red, 2pt]{\cases{a_0=\pi \\ a_1=4/\pi}}$$

解答:$$L\{y''\} -4L\{y'\} =2L\{\delta(t)\} \Rightarrow s^2Y(s)-4sY(s)=2 \Rightarrow Y(s)= {2\over s^2-4s} \\ \Rightarrow y(t)=L^{-1}\{ Y(s)\} =L^{-1}\{ {2\over s^2-4s}\}= L^{-1}\{ {1/2\over s-4} -{1/2\over s}\}={1\over 2}e^{4t}-{1\over 2}\\ \Rightarrow \bbox[red, 2pt]{y(t)= {1\over 2}e^{4t}-{1\over 2}}$$

解答:$$y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1=0 \Rightarrow (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow y_h=c_1e^t +c_2xe^t\\ \text{Apply variation of parameters, let }\cases{y_1=e^t \\ y_2=te^t\\ r(t) =e^t} \\\Rightarrow W= \begin{vmatrix}y_1 & y_2 \\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix}e^t & te^t \\ e^t & e^t+te^t \end{vmatrix}= e^{2t} \Rightarrow y_p = -y_1\int {y_2\cdot r(t)\over W}\,dt +y_2 \int {y_1\cdot r(t)\over W}\,dt \\ \Rightarrow y_p = -e^t \int {te^{2t}\over e^{2t}} \,dt + te^t \int 1\,dt =-{1\over 2}e^t t^2+ t^2e^t= {1\over 2}t^2e^t \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y(t)=c_1e^t+ c_2te^t+ {1\over 2}t^2e^t}$$

==================== END ======================

解題僅供參考,其他歷年試題及詳解