113年專門職業及技術人員高等考試
等 別：高等考試
類 科：電子工程技師
科 目：工程數學（包括線性代數、微分方程、向量分析、複變函數與機率）
解答:$$\textbf{(一) }\frac{d^5 v}{dt^5}+v=0 \Rightarrow \lambda^5+1=0 \Rightarrow \lambda=e^{i\pi /5},e^{i 3\pi/5}, e^{i \pi}, e^{i 7\pi /5}, e^{i 9\pi/5} =\sigma_1\pm i\omega_1, \sigma_2\pm i\omega_2,-1\\ \quad \Rightarrow \bbox[red, 2pt]{v_h=e^{\sigma_1 t}(c_1 \cos(\omega_1 t) +c_2 \sin(\omega_1 t)) + e^{\sigma_2 t}(c_3 \cos(\omega_2 t) +c_4 \sin(\omega_2 t)) + c_5e^{-t}} \\\textbf{(二) }L\left\{ {d^5 v\over dt^5} \right\} +L\{v\}=L\left\{ e^{-2t}\right\} \Rightarrow s^5V(s)-s^4 +V(s)={1\over s+2} \Rightarrow V(s)={1\over (s+2)(s^5+1)}+ {s^4\over s^5+1} \\ \text{Let }\lambda_k =e^{i(\pi+2k\pi)/5}(k=0,...,4) \text{ denote the five roots of }s^5+1=0\\ {s^4\over s^5+1} \text{ has simple pole }s=\lambda_k \Rightarrow \text{Res}_{s=\lambda_k}{s^4\over s^5+1} = \left. {s^4\over (s^5+1)'}\right|_{s= \lambda_k} ={\lambda_k^4\over 5\lambda_k^4} ={1\over 5}\\\qquad \Rightarrow L^{-1} \left\{{s^4\over s^5+1} \right\} =\sum_{k=0}^4 {1\over 5}e^{\lambda_k t}\\ {1\over (s+2)(s^5+1)} \text{ has simples at s=-2 and }s=\lambda_k \Rightarrow \text{Res}_{s=-2}{1\over (s+2)(s^5+1)} = -{1\over 31} \\ \text{Res}_{s=\lambda_k}{1\over (s+2)(s^5+1)} = \left. {1\over (s+2)(s^5+1)'}\right|_{s=\lambda_k} ={1\over (\lambda_k+2)5\lambda_k^4} \\ \Rightarrow L^{-1} \left\{ {1\over (s+2)(s^5+1)}\right\} =-{1\over 31} e^{-2t} + \sum_{k=0}^4 {1\over (\lambda_k+2)5 \lambda_k^4} e^{\lambda_k t} \\ \Rightarrow v(t) =L^{-1}\{V(s)\} =\sum_{k=0}^4 {1\over 5}e^{\lambda_k t}-{1\over 31} e^{-2t} + \sum_{k=0}^4 {1\over (\lambda_k+2)5 \lambda_k^4} e^{\lambda_k t} \\=-{1\over 31}e^{-2t} + \sum_{k=0}^4 e^{\lambda_k t} \left( {1\over 5}+{1\over (\lambda_k+2)5 \lambda_k^4} \right) =-{1\over 31}e^{-2t} + \sum_{k=0}^4 e^{\lambda_k t} \left( {2\over 5(\lambda_k+2)} \right) \\ \Rightarrow \bbox[red, 2pt]{v(t)=-{1\over 31}e^{-2t} +{2\over 5}e^{-t}+{ 4\over 5(5+4\sigma_1)}e^{\sigma_1 t} \left( (\sigma_1+2) \cos(\omega_1 t)+ \omega_1 \sin(\omega_1 t) \right)} \\\qquad \bbox[red, 2pt]{+ {4\over 5(5+ 4\sigma_2)} e^{\sigma_2 t} \left( (\sigma_2+2) \cos(\omega_2 t) + \omega_2 \sin(\omega_2 t) \right)}$$
解答:$$\textbf{(一) }Q= \begin{bmatrix} -4 & 1 & 0 \\1 & -7 & 3 \\0 & 3 & -4\end{bmatrix} \Rightarrow \det(Q-\lambda I) =-\lambda^3-15\lambda^2-62\lambda-72 =-(\lambda+2) (\lambda+9)(\lambda+4) \\ \quad \Rightarrow \lambda=-2,-9,-4\\ \lambda_1=-2 \Rightarrow (Q-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} -2 & 1 & 0 \\1 & -5 & 3 \\ 0 & 3 & -2\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3/3\\ x_2=2x_3/3} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}1/3\\ 2/3\\ 1 \end{bmatrix}, \text{choose }v_1= \begin{bmatrix}1/3\\ 2/3\\ 1 \end{bmatrix} \\ \lambda_2= -9 \Rightarrow (Q-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} 5 & 1 & 0 \\1 & 2 & 3 \\0 & 3 & 5\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3/3\\ x_2=-5x_3/3} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}1/3\\ -5/3\\ 1 \end{bmatrix}, \text{choose }v_2 =\begin{bmatrix}1/3\\ -5/3\\ 1 \end{bmatrix} \\ \lambda_3= -4 \Rightarrow (Q-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} 0 & 1 & 0 \\1 & -3 & 3 \\0 & 3 & 0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1= -3x_3\\ x_2=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}-3\\ 0\\ 1 \end{bmatrix}, \text{choose }v_3 = \begin{bmatrix}-3\\ 0\\ 1 \end{bmatrix} \\\quad \Rightarrow \bbox[red, 2pt]{特徵值:-2,-9,-4; 對應之特徵向量:\begin{bmatrix}1/3\\ 2/3\\ 1 \end{bmatrix}, \begin{bmatrix}1/3\\ -5/3\\ 1 \end{bmatrix}, \begin{bmatrix}-3\\ 0\\ 1 \end{bmatrix}} \\\textbf{(二) }所有特徵值皆為負值，因此矩陣Q為\bbox[red, 2pt]{\text{negative definite}}\\ \quad 根據\text{Rayleigh Quotient},極大值即為最大的特徵值;極小值即為最小的特徵值 \Rightarrow \bbox[red, 2pt]{\cases{極大值-2\\ 極小值-9}}$$
解答:$$\textbf{(一) }穩態時，即{\partial T(x,t) \over \partial t} =0 \Rightarrow 4{\partial^2 T(x,t) \over \partial x^2} =0 \Rightarrow 穩定狀態T(x,t)=T_s(x) =Ax+B \\ \quad \Rightarrow {\partial T(x,t) \over \partial x} =A \Rightarrow 邊界條件:\cases{ T(0,t)=B=0\\ T_x(1,t)=A=1} \Rightarrow \bbox[red, 2pt]{T_s(x)=x, 0\le x\le 1} \\\textbf{(二) }T(x,t)=T_s(x)+u(x,t) \Rightarrow T_t=4T_{xx} \Rightarrow u_t=4u_{xx} \\ \quad 邊界條件:\cases{T(0,t)=T_s(0)+u(0,t) =u(0,t)=0\\ T_x |_{x=1} \Rightarrow T'_s(1)+ u_x(1,t)=1 \Rightarrow u_x(1,t) =0}\\ 分離變數: u(x,t)=X(x)T(t) \Rightarrow u_t=4u_{xx} \Rightarrow {T' \over 4T} ={X'' \over X}=k \\ k=-\mu^2 \Rightarrow X''+\mu^2 X=0 \Rightarrow X= A\cos(\mu x)+B\sin (\mu x) \Rightarrow X'=-A\mu \sin(\mu x)+B\mu \cos (\mu x) \\ \quad \Rightarrow \cases{X(0)=A=0\\ X'(1) =B\mu \cos(\mu) =0} \Rightarrow \mu ={(2n+1)\pi\over 2},n=0,1,2,... \\ \Rightarrow 特徵函數: \bbox[red, 2pt]{X_n(x) =\sin{(2n+1)\pi x\over 2},n=0,1,2,\dots}$$
解答:$$\textbf{(一) } \cases{x(t) = \cos t\\ y(t)= \sin t} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t) = \cos t} \Rightarrow \cases{F_x=4\cos t+3\sin t\\ F_y=\cos t+ 2\sin t} \\ \Rightarrow W= \int_C F\cdot dr = \int_0^\pi \left[ (4\cos t+ 3 \sin t)(-\sin t)+( \cos t+2 \sin t)(\cos t) \right]\,dt \\\quad = \int_0^\pi (-1+ 2\cos(2t)-\sin(2t))\,dt = \bbox[red, 2pt]{-\pi} \\\textbf{(二) } \oint_D F\cdot dr = \iint _D \left( {\partial F_y\over \partial x} -{\partial F_x\over \partial y} \right) dA = \iint_D(1-3)dA =-2\times 半圓面積= \bbox[red, 2pt]{-\pi}$$
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