111年北科大電機工程碩士班-工程數學詳解

國立 臺北科 技 大學 111學 年度碩 士班招 生考試

系所組別 :2131電 機工程系碩士班丙組
第一節 工程數學 試題 (選考)

解答： $$\textbf{(a)}\;令\cases{P(x,y)=\cos(x+y)-\sin(x+y)\\ Q(x,y)=-\sin(x+y)} \Rightarrow \cases{\mu P=(\cos(x+y)-\sin(x+y))e^{ax} \\ \mu Q= -\sin(x+y)e^{ax}} \\\qquad \Rightarrow \cases{(\mu P)_y=-(\cos(x+y) +\sin(x+y))e^{ax}\\ (\mu Q)_x= -(\cos(x+y)+ a\sin(x+y))e^{ax}} \\ \qquad \mu為積分因子\Rightarrow (\mu P)_y=(\mu Q)_x \Rightarrow \bbox[red, 2pt]{a=1} \\\textbf{(b)}\; \Phi(x,y)=\int \mu P\,dx = \int \mu Q\,dy \\\qquad \Rightarrow \int (\cos(x+y)-\sin(x+y))e^{x}\,dx = \int -\sin(x+y)e^{x}\,dy \\\qquad \Rightarrow \Phi(x,y)=e^x\cos(x+y)+ \rho(y)=e^x\cos(x+y)+ \phi(x) \\ \qquad\Rightarrow \text{ the general solution: }\bbox[red,2pt]{e^x\cos(x+y)=c_1}$$

解答： $$y=(x+5)v(x) \Rightarrow y'=v+ (x+5)v' \Rightarrow y''=2v'+ (x+5)v'' \\ 原式變為 (x+5)^2(2v'+ (x+5)v'') -3(x+5)(v+(x+5)v') +3(x+5)v=x \\ \Rightarrow (x+5)^3v''-(x+5)^2 v' =x \Rightarrow {1\over x+5}v''-{1\over (x+5)^2}v'={x \over (x+5)^4} \\ \Rightarrow \left( {1\over x+5}v' \right)'={x \over (x+5)^4} \Rightarrow {1\over x+5} v'=\int {x \over (x+5)^4}\,dx ={3x+5\over 6(x+5)^3} +c_1\\ \Rightarrow v'= {3x+5\over 6(x+5)^2} + c_1(x +5)  \Rightarrow v= \int \left( {3x+5\over 6(x+5)^2}+c_1(x+5) \right) \,dx \\={5\over 3(x+5)}+ {1\over 2}\ln(x+5) +{1\over 2}c_1(x+5)^2+c_2\\ \Rightarrow y=(x+5)v=  {5\over 3}+ {1\over 2}(x+5) \ln(x+5) +{1\over 2}c_1(x+5)^3+c_2(x+5) \\ \Rightarrow \bbox[red, 2pt]{y={5\over 3}+ {1\over 2}(x+5) \ln(x+5) + c_3(x+5)^3+c_2(x+5)}$$

解答： $$F''(s)+ 5F'(s)+ 6F(s)=0 \Rightarrow F(s)= c_1e^{3s}+ c_2 e^{2x} \Rightarrow F'(s)= 3c_1e^{3s}+2c_2e^{2s} \\ \Rightarrow \cases{F(0)= c_1+ c_2=1\\ F'(0)= 3c_1+ 2c_2=0} \Rightarrow \cases{c_1=-2\\ c_2=3} \Rightarrow F(s)=-2e^{3s}+ 3e^{2s}\\ \Rightarrow f(t)= L^{-1}\{F(s)\} =-2L^{-1}\{e^{3s}\}+3 L^{-1}\{ e^{2s}\} \Rightarrow \bbox[red, 2pt]{f(t) =-2\delta(t+3) +3\delta(t+2)}$$

解答： $$\textbf{(a)}\; \cases{取E_1=\left( \begin{matrix}1 & 0 & 0 \\-3 & 1 & 0 \\-3 & 0 & 1\end{matrix} \right) \\[1ex] A=\left(\begin{matrix} 1 & 2 & 3 \\ 3 & 5 & 1 \\3 & 5 & 7 \end{matrix} \right)} \Rightarrow E_1A=\left(\begin{matrix} 1 & 2 & 3 \\0 & -1 & -8 \\0 & -1 & -2\end{matrix} \right)\\ E_2=\left( \begin{matrix}1 & 2 & 0 \\0 & 1 & 0 \\0 & -1 & 1\end{matrix} \right) \Rightarrow E_2E_1A= \left( \begin{matrix} 1 & 0 & -13 \\0 & -1 & -8 \\0 & 0 & 6\end{matrix} \right) \\ 又E_3=\left(\begin{matrix} 1 & 0 & \frac{13}{6} \\0 & 1 & \frac{8}{6} \\0 & 0 & 1\end{matrix} \right) \Rightarrow E_3E_2E_1A= \left( \begin{matrix}1 & 0 & 0 \\0 & -1 & 0 \\0 & 0 & 6 \end{matrix} \right)\\ 因此取U= E_3E_2E_1 \Rightarrow \bbox[red, 2pt]{U= \left(\begin{matrix} 1 & \frac{-1}{6} & \frac{13}{6} \\ -3 & \frac{1}{6} & \frac{-31}{6} \\-3 & \frac{-1}{2} & \frac{-11}{2} \end{matrix} \right), V=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 &1 \end{matrix} \right)}\\ \textbf{(b)}\; \cases{E_1^{-1} =\left(\begin{matrix} 1 & 0 & 0 \\3 & 1 & 0 \\3 & 0 & 1\end{matrix}\right) \\[1ex] E_2^{-1} = \left(\begin{matrix}1 & -2 & 0 \\0 & 1 & 0 \\0 & 1 & 1\end{matrix} \right) \\[1ex] \\ E_3^{-1} = \left( \begin{matrix} 1 & 0 & \frac{-13}{6} \\0 & 1 & \frac{-8}{6} \\0 & 0 & 1 \end{matrix} \right)} \Rightarrow U^{-1}=E_1^{-1}E_2^{-1}E_3^{-1} \Rightarrow \bbox[red, 2pt]{U^{-1}= \left(\begin{matrix} \frac{-7}{2} & -2 & \frac{1}{2} \\-1 & 1 & \frac{-4}{3} \\2 & 1 & \frac{-1}{3} \end{matrix} \right)} \\ V=I_3 \Rightarrow \bbox[red, 2pt]{V^{-1} =\begin{pmatrix}1& 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1 \end{pmatrix}}$$

解答： $$\lambda^3-4\lambda^2+5 \lambda-2=(\lambda-1)^2(\lambda-2) =0 \Rightarrow \text{eigen values: }\lambda_1=1,\lambda_2=2 \\ f(\lambda)= c_0+ c_1\lambda +c_2\lambda^2 \Rightarrow f'(\lambda)= c_1+2c_2\lambda \\令e^A=c_0I+c_1A+ c_2A^2 \Rightarrow \cases{e^1=c_0+c_1\\ e^2=c_0+2c_1+4c_2\\ e^1= c_1+2c_2} \Rightarrow \cases{c_0=e^2-2e\\ c_1=-e^2+3e\\ c_2=e^2/2-e} \\ \Rightarrow \bbox[red, 2pt]{e^A= \left( {e^2 \over 2}-e\right)A^2+ (-e^2+2e)A+e^2-2e}$$

解答： $$\textbf{(a)}\; A= \begin{bmatrix}1 & 0& -2& 1 \\0 & -1 & 1& 0 \end{bmatrix}, \mathbf{x} =\begin{bmatrix}x_1 \\x_2\\ x_3\\x_4 \end{bmatrix} \Rightarrow A \ \mathbf x=0 \Rightarrow \cases{x_1-2x_3+x_4=0 \\ -x_2+x_3=0} \\ \quad \Rightarrow \cases{x_1+x_4= 2x_3 \\x_2=x_3} \Rightarrow \mathbf{x}=x_3 \begin{bmatrix}-2 \\1\\ 1\\0 \end{bmatrix} + x_4 \begin{bmatrix}-1 \\0\\ 0\\1 \end{bmatrix}\\\quad   \Rightarrow \bbox[red, 2pt]{Null(A)= \left\{ a \begin{pmatrix}-2 \\1\\ 1\\0 \end{pmatrix} + b \begin{pmatrix}-1 \\0\\ 0\\1 \end{pmatrix} \mid a,b\in \mathbb R\right\}} \\\textbf{(b)}\; \text{Let }u=(a,b,c,d)^T \text{ be a component of the orthogonal complement of }N(A). \\ \Rightarrow \cases{u\cdot (-2,1,1,0)=0\\ u \cdot (-1,0,0,1)=0} \Rightarrow \cases{2a=b+c \\ a+d=0} \Rightarrow u= a\begin{pmatrix} 1\\ 0 \\2 \\ 1 \end{pmatrix} +b\begin{pmatrix} 0\\ 1 \\-1 \\ 0\end{pmatrix} \\ \Rightarrow \text{orthogonal complement of }N(A) = \bbox[red, 2pt]{W^\bot =\left\{ a\begin{pmatrix} 1\\ 0 \\2 \\ 1 \end{pmatrix} +b\begin{pmatrix} 0\\ 1 \\-1 \\ 0\end{pmatrix} \mid a,b\in \mathbb R\right\}}\\ \textbf{(c)}\; \text{It is clear that: } \vec 0\in W^\bot. \text{If }u \in W^\bot \text{ and }v\in W^\bot, \text{ then }au+bv \in W^\bot, \forall a,b\in \mathbb R\\ \qquad \text{That is, }W^\bot \text{is a subspace of }R^4. \bbox[red, 2pt]{Q.E.D.}$$
 

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解題僅供參考, 其它歷年試題及詳解