113年雲科大電子碩士班-工程數學詳解

國立雲林科技大學113學年度碩士班招生考試

系所：電子系
科目：工程數學 

解答：

解答： $$\cases{P(x,y)= 3xy-y^2\\ Q(x,y)=x(x-y)} \Rightarrow \cases{P_y=3x-2y \\ Q_x = 2x-y} \Rightarrow P_y \ne Q_x \Rightarrow \text{ Not Exact} \\ {P_y-Q_x\over Q} ={x-y\over x(x-y)} ={1\over x} \Rightarrow u'={1\over x} u \Rightarrow \text{integrating factor }u(x)=x \\ \Rightarrow \cases{uP= 3x^2y-xy^2\\ uQ=x^2(x-y)} \Rightarrow (uP)_y=3x^2-2xy= (uQ)_x \Rightarrow \text{Exact } \\ \Rightarrow \Phi(x,y)= \int (3x^2y-xy^2)\,dx = \int x^2(x-y)\, dy \\ \Rightarrow \Phi=x^3y-{1\over 2}x^2y^2+ \phi(y) = x^3y-{1\over 2}x^2y^2 + \rho(x) \\ \Rightarrow \bbox[red, 2pt]{x^3y-{1\over 2}x^2y^2+c_1=0}$$

解答： $$y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1=0 \Rightarrow (\lambda-1)^2 =0 \Rightarrow \lambda=1 \Rightarrow y_h= c_1e^x+ c_2 xe^x\\ \text{Applying variations of parameters, let} \cases{y_1=e^x\\ y_2=xe^x} \Rightarrow W=\begin{vmatrix}y_1 & y_2 \\y_1' & y_2' \end{vmatrix} =e^{2x} \\ \Rightarrow y_p = -e^x \int{xe^x \cdot {e^x \over x^2} \over e^{2x}}\,dx + xe^x \int { e^x \cdot {e^x\over x^2}\over e^{2x}} \,dx =-e^x \int {1\over x}\,dx +xe^x \int {1\over x^2}\,dx \\\qquad =-e^x \ln x-e^x \\ \Rightarrow y= y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_3e^x+c_2xe^x-e^x\ln x}$$

解答： $$\textbf{(a)}\; \cases{y_1'=y_1+2y_2 \cdots(1) \\ y_2'= 3y_1+2y_2 \cdots(2)} \Rightarrow y_1''=y_1'+2y_2' =y_1'+2(3y_1+2y_2)=y_1'+6y_1+4y_2 \\ \Rightarrow y_1''-y_1'-6y_1=4y_2=2(y_1'-y_1) =2y_1'-2y_1 \Rightarrow y_1''-3y_1'-4y_1=0 \\ \Rightarrow y_1=c_1 e^{4t}+c_2e^{-t} \Rightarrow y_1'=4c_1e^{4t}-c_2e^{-t} \Rightarrow 2y_2=y_1'-y_1=3c_1e^{4t}-2c_2e^{-t} \\ \Rightarrow y_2={3\over 2}c_1e^{4t}-c_2e^{-t} \Rightarrow \bbox[red, 2pt]{\cases{y_1= c_1 e^{4t}+c_2e^{-t}\\ y_2={3\over 2}c_1e^{4t}-c_2e^{-t}}}$$

解答： $$\textbf{(a)}\; \cases{A= \begin{bmatrix}1 & 0 & -1\\2 & 3 & 1 \end{bmatrix} \\[1ex] B=\begin{bmatrix}1 & 2 & 0\\0 & 4 & 3 \end{bmatrix}} \Rightarrow A+B = \begin{bmatrix}2 & 2 & -1\\2 & 7 & 4 \end{bmatrix} \Rightarrow (A+B)^T = \bbox[red, 2pt]{\begin{bmatrix}2 & 2\\2 & 7\\-1 & 4 \end{bmatrix} } \\\textbf{(b)}\; 2A-B=\begin{bmatrix}1 & -2 & -2\\4 & 2 & -1 \end{bmatrix} \Rightarrow (2A-B)C = \bbox[red, 2pt]{ \begin{bmatrix}-2 & -1\\-4 & 3\end{bmatrix}} \\\textbf{(c)}\; A-B=\begin{bmatrix}0 & -2 & -1\\2 & -1 & -2 \end{bmatrix} \Rightarrow 4(A-B)= \begin{bmatrix} 0 & -8 & -4\\8 & -4 & -8 \end{bmatrix} \\\qquad 2X+4(A-B)=0 \Rightarrow 2X=-4(A-B) =\begin{bmatrix}0 & 8 & 4\\-8 & 4 & 8 \end{bmatrix} \Rightarrow X= \bbox[red, 2pt]{\begin{bmatrix}0 & 4 & 2\\-4 & 2 & 4 \end{bmatrix}}$$

解答： $$A=\begin{bmatrix}2 & 1 & 1 \\1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -(\lambda-1)^2(\lambda-4)=0 \Rightarrow \lambda=1,4\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0 \Rightarrow x_1+x_2+x_3=0\\ \qquad \Rightarrow v=x_2 \begin{pmatrix}-1 \\1\\ 0 \end{pmatrix} +x_3  \begin{pmatrix}-1 \\0\\ 1 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix}-1 \\1\\ 0 \end{pmatrix}, v_2= \begin{pmatrix}-1 \\0\\1 \end{pmatrix} \\ \lambda_2=4 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2 & 1 & 1 \\ 1 & -2 & 1 \\1 & 1 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \\ \qquad \Rightarrow v=x_3 \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix} \text{ choose }v_3= \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}\\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{1,4}, \text{ and eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix}-1 \\1\\ 0 \end{pmatrix},\begin{pmatrix}-1 \\0\\1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}}$$

解答： $$\textbf{(a)}\; \vec A\cdot \vec B = (2,1,-1)\cdot (1,3,-1)= 2+3+1=\bbox[red, 2pt] 6\\ \textbf{(b)}\; \vec A\times \vec B=\begin{vmatrix}\vec i & \vec j&\vec k \\2 & 1 & -1\\ 1& 3 & -1 \end{vmatrix} =\bbox[red, 2pt]{2\vec i+1\vec j+5\vec k} \\\textbf{(c)}\; {\vec A\cdot \vec B\over |\vec B|} ={6\over \sqrt{11}} =\bbox[red, 2pt]{6\sqrt{11}\over 11}$$

解答： $$\text{Let }f(x,y,z)=2(x^2+y^2 )-z^2 \Rightarrow \nabla f=(4x,4y,-2z) \Rightarrow \nabla f(1,0,1)= \bbox[red, 2pt]{(4,0,-2)}$$

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