111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：$$先求齊次解，即y''+y'-2y=0 \Rightarrow y_h= C_1e^{-2t} +C_2e^t\\ \text{Case I }(0\lt t\lt 2\pi): y_p = A\sin t+B\cos t \Rightarrow y_p' = A\cos t-B\sin t \Rightarrow y_p''=-A\sin t-B\cos t\\ \Rightarrow y_p''+ y_p'-2y_p = (-3A-B)\sin t+(A-3B) \cos t=3\sin t-\cos t \Rightarrow \cases{-3A-B=3\\ A-3B= -1} \\ \Rightarrow \cases{A=-1\\ B=0} \Rightarrow y_p =-\sin t \Rightarrow y=y_h+ y_p = C_1e^{-2t} +C_2e^t-\sin t \\\Rightarrow y'=-2C_1e^{-2t} +C_2e^t- \cos t \Rightarrow \cases{y(0)= 1 = C_1+ C_2 \\ y'(0)= 0 = -2C_1 +C_2-1} \Rightarrow \cases{C_1=0\\ C_2=1} \Rightarrow y=e^t-\sin t \\ \text{Case II }(t \gt 2\pi): y_p = A\sin (2t)+B\cos (2t) \Rightarrow y_p' = 2A\cos(2t) -2B\sin(2t) \\ \Rightarrow y_p'' = -4A \sin(2t)-4B\cos(2t) \Rightarrow y_p''+ y_p'-2y_p = (-6A-2B)\sin(2t) +(2A-6B)\cos (2t) \\= 3\sin(2t)-\cos(2t) \Rightarrow \cases{-6A-2B=3 \\ 2A-6B= -1} \Rightarrow \cases{A=-1/2\\ B=0 } \Rightarrow y_p = -{1\over 2}\sin (2t) \\ \Rightarrow y=y_h+ y_p = C_1e^{-2t} +C_2e^t-{1\over 2}\sin(2t) \Rightarrow y'=-2C_1e^{-2t}+ C_2e^t-\cos(2t)\\ \Rightarrow \cases{y(0)= 1 = C_1+C_2 \\ y'(0)=0 = -2C_1+C_2-1} \Rightarrow \cases{C_1=0 \\C_2= 1} \Rightarrow y= e^t-{1\over 2}\sin(2t) \\ \Rightarrow \bbox[red, 2pt]{\begin{cases}y=e^t-\sin t,& 0\lt t\lt 2\pi \\ y=e^t-{1\over 2}\sin(2t), & t\gt 2\pi \end{cases}}$$
解答：$$\cases{A=\begin{bmatrix} 1 & 2 & 3 & 4\\ 1& 1& 1& 1\end{bmatrix} \\ AB=I_2} \Rightarrow B= \begin{bmatrix} x_1 & x_5 \\ x_2 & x_6\\ x_3 &x_7\\ x_4 & x_8\end{bmatrix}\\ \left[\begin{array}{cccc|cc} 1 & 2 & 3 & 4 & 1 & 0\\ 1& 1& 1& 1 & 0 & 1\end{array} \right] \xrightarrow{-r_1+r_2 \to r_2} \left[\begin{array}{cccc|cc} 1 & 2 & 3 & 4 & 1 & 0\\ 0& -1& -2 & -3 & -1 & 1\end{array} \right]\\ \xrightarrow{2r_2+r_1 \to r_1}\left[\begin{array}{cccc|cc} 1 & 0 & -1 & -2 & -1 & 2\\ 0& -1& -2 & -3 & -1 & 1\end{array} \right] \xrightarrow{-r_2} \left[\begin{array}{cccc|cc} 1 & 0 & -1 & -2 & -1 & 2\\ 0& 1& 2 & 3 & 1 & -1\end{array} \right]\\ \Rightarrow \cases{\cases{x_1-x_3-2x_4=-1\\ x_2+2x_3 +3_4=1} \Rightarrow 取 (x_1,x_2,x_3,x_4) =(1,-3,2,0) \\ \cases{x_5-x_7-2x_8=2\\ x_6+2x_7+ 3x_8 =-1} \Rightarrow 取 (x_5,x_6,x_7, x_8)= (1,1,-1,0)} \\ \Rightarrow \bbox[red, 2pt]{B=\begin{bmatrix} 1 & 1\\ -3 & 1 \\ 2& -1 \\ 0 & 0\end{bmatrix}}$$
解答：$$\cases{x(t)= \cos t\\ y(t)=\sin t\\ z(t)=t/3} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t) = \cos t\\ z'(t)= 1/3} \Rightarrow 曲線長= \int_{-4\pi}^{4\pi} \sqrt{x'(t)^2 +y'(t)^2 +z'(t)^2} \,dt \\=\int_{-4\pi}^{4\pi} \sqrt{\sin^2 t + \cos^2 t +1/9} \,dt =\int_{-4\pi}^{4\pi} \sqrt{1+1/9} \,dt =\int_{-4\pi}^{4\pi} \sqrt{10/9} \,dt =\sqrt{10\over 9} \times 8\pi = \bbox[red, 2pt]{{8\over 3} \sqrt{10}\pi}$$
解答：$$\mathbf{(一)}\;令f(z)={(z+1)\sin z\over 4} \Rightarrow f'(z)= {\sin z+(z+1)\cos z\over 4} \Rightarrow f'(1/2)= {\sin(1/2) + (3/2)\cos (1/2) \over 4} \\ \quad\Rightarrow \oint_C {(z+1)\sin z\over (2z-1)^2}\,dz = \oint {f(z)\over (z-1/2)^2 }\,dz = 2\pi i\cdot f'(1/2) =\bbox[red, 2pt]{\pi i\left({1\over 2}\sin {1\over 2}+ {3\over 4}\cos {1\over 2} \right)} \\\mathbf{(二)}\;令f(z)= \cos z \Rightarrow f'(z)=-\sin z \Rightarrow f''(z)=-\cos z \Rightarrow f'''(z)= \sin z \Rightarrow f^{[4]}(z)= \cos z\\ \quad \Rightarrow f^{[5]}(z) = -\sin z \Rightarrow f^{[6]}(z) = -\cos z \Rightarrow \oint {\cos z\over z^7}\,dz = {2\pi i \over 6!} \cdot f^{[6]}(0) = \bbox[red, 2pt]{-{\pi \over 360}}$$
解答：$$\cases{\sigma_X^2 = E[X^2]-(E[X])^2 \\ \sigma_Y^2 = E[Y^2]-(E[Y])^2 } \Rightarrow \cases{4= E[X^2]-0 \\ 16= E[Y^2]-0} \Rightarrow \cases{E[X^2]=4\\ E[Y^2]= 16}\\ 又相關係數\rho_{XY}={Cov(X,Y)\over \sigma_X \cdot \sigma_Y} \Rightarrow Cov(X,Y) = -0.25\cdot 2 \cdot 4= -2 =E[XY]-E[X]\cdot E[Y]\\ \Rightarrow E[XY]= -2-0 \Rightarrow E[XY]= -2\\ 因此 E[W]=E[(aX+ 3Y)^2] =E[a^2X^2 +6aXY +9Y^2] = a^2E[X^2]+ 6aE[XY]+ 9E[Y^2] \\ =4a^2 -12a + 144 \Rightarrow 當a={12\over 8}= {3\over 2}時，E[W]=135為最小值;\\ 答：\cases{\mathbf{(一)} \bbox[red,2pt]{a=3/2} \\ \mathbf{(二)} 最小值\bbox[red, 2pt]{E[W]=135}}$$
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