110年北科大光電碩士班-工程數學詳解

 國立臺北科技大學110學年度碩士班招生考試

系所組別:光電工程碩士班

科目:工程數學

解答: $$y' =\frac{\text{d}y}{\text{d}x} =-{y+y^2 \over x(1+2y)} \Rightarrow {1+2y\over y+y^2}dy=-{1\over x}dx \Rightarrow \ln y+\ln(y+1)=-\ln x+c_1 \\ \Rightarrow \ln y(y+1)=\ln {c_2\over x} \Rightarrow y(y+1)={c_2\over x} \Rightarrow \bbox[red, 2pt]{y={-1\pm \sqrt{1+4c_2/x}\over 2}}$$

解答: $$y''+2y'+y=0 \Rightarrow \lambda^2+2\lambda +1=0 \Rightarrow (\lambda+1)^2=0 \Rightarrow \lambda=-1 \Rightarrow y_h= c_1e^{-x} +c_2xe^{-x}\\ \text{Applying variation of parameters, let }\cases{y_1=e^{-x}\\ y_2= xe^{-x}} \Rightarrow W=\begin{vmatrix}y_1 & y_2 \\y_1' & y_2' \end{vmatrix} =\begin{vmatrix}e^{-x} & xe^{-x} \\-e^{-x} & e^{-x}-xe^{-x} \end{vmatrix} =e^{-2x}\\ y_p = -e^{-x} \int{ xe^{-x} (4e^{-x}\ln x) \over e^{-2x}}\,dx + xe^{-x} \int { e^{-x}(4e^{-x}\ln x)\over e^{-2x}}\,dx \\=-e^{-x} \int 4x\ln x\,dx +xe^{-x} \int 4\ln x\,dx =-4e^{-x}\left( {1\over 2}x^2\ln x-{1\over 4}x^2\right) +4xe^{-x}(x\ln x-x) \\=2x^2e^{-x} \ln x-3x^2e^{-x} \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1e^{-x} +c_2xe^{-x}+ 2x^2e^{-x} \ln x-3x^2e^{-x}}$$

解答: $$\textbf{(a)}\; A=\begin{bmatrix}\cos \theta & 0 & \sin \theta \\0 & 1& 0 \\ -\sin \theta & 0 &\cos \theta\end{bmatrix} \text{ is a rotation matrix, then } A^{-1} =A^T\\ \qquad \Rightarrow A^{-1}  = \bbox[red, 2pt]{\begin{bmatrix} \cos \theta & 0 & -\sin \theta \\0 & 1& 0 \\ \sin \theta & 0 &\cos \theta \end{bmatrix} } \\ \textbf{(b)}\; \det(A)= 1 \ne 0\Rightarrow \bbox[red, 2pt]{非奇異}$$

解答: $$A \xrightarrow{R_2-4R_3\to R_2} \begin{bmatrix}a_3 & a_2 & a_1 \\b_3 & b_2& b_1 \\ c_3& c_2 & c_1 \end{bmatrix} \Rightarrow \det(A)= \bbox[red, 2pt]{-5}$$

解答: $$\textbf{(a)}\; \text{div }\vec V =\frac{\partial \vec V_1} {\partial x} +\frac{\partial \vec V_2}{\partial y} +\frac{\partial \vec V_3}{\partial z} = 2xy+ 6yz+ 2zx \\\quad \Rightarrow \text{div }\vec V(1,0,1)= \bbox[red, 2pt]2 \\\textbf{(b)}\; 流場在某點的散度代表該點為匯聚點或發源點及其流量強度$$

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解題僅供參考,其他歷年試題及詳解