114年台北科大電機碩士班-工程數學詳解

國立臺北科技大學114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
第二節 工程數學 試題(選考)

解答: $$v=y^{-3/2} \Rightarrow v'=-{3\over 2}y^{-5/2}y' \Rightarrow y'=-{2\over 3}y^{5/2}v' \Rightarrow -{2\over 3} xy^{5/2}v'+y=(xy)^{5/2} \\ \Rightarrow -{2\over 3}xv'+v=x^{5/2}   \Rightarrow x^{-3/2}v'-{3\over 2}x^{-5/2}=-{3\over 2} \Rightarrow \left( x^{-3/2}v\right)' =-{3\over 2} \Rightarrow x^{-3/2}v=-{3\over 2}x+c_1 \\ \Rightarrow v=y^{-3/2} =-{3\over 2}x^{5/2} +c_1x^{3/2} \Rightarrow \bbox[red, 2pt]{y= \left( -{3\over 2}x^{5/2}+c_1x^{3/2} \right)^{-2/3}}$$

解答: $$L\{y''\}+ 4L\{y'\}+3 L\{y\} =L\{\delta(t-\pi)\} \Rightarrow s^2Y(s)+4sY(s)+3Y(s)= e^{-\pi s} \\ \Rightarrow Y(s) ={e^{-\pi s} \over s^2+4s+3} \Rightarrow y(t)=L^{-1}\{Y(s)\} =L^{-1} \left\{ {e^{-\pi s} \over s^2+4s+3}\right\} \\ =L^{-1} \left\{ {1\over 2}e^{-\pi s}\left( { 1\over s+1} -{1\over s+3}\right)\right\}   \Rightarrow \bbox[red, 2pt]{y(t)= {1\over 2}u(t-\pi)\left( e^{-(t-\pi)}-e^{-3(t-\pi)}\right)}$$

解答: $$1. \begin{bmatrix}0&0 \\0& 0 \end{bmatrix} \in U\\ 2. \text{ If }U_1=\begin{bmatrix}a&b \\c& d \end{bmatrix} \in U, \text{ then }a+b+c+d=0 \Rightarrow kU_1= \begin{bmatrix}ka& kb \\kc& kd \end{bmatrix} \Rightarrow k(a+b+c+d)=0 \Rightarrow kU_1\in U \\ 3. \text{ If }\cases{U_1 =\begin{bmatrix} a_1 &b_1 \\c_1& d_1 \end{bmatrix} \in U \\U_2=\begin{bmatrix}a_2 &b_2 \\c_2& d_2 \end{bmatrix} \in U},\text{ then }\cases{a_1+b_1 +c_1+ d_1=0 \\ a_2+b_2+c_2 +d_2=0} \Rightarrow U_1+U_2 =\begin{bmatrix} a_1+a_2 &b_1+b_2 \\c_1+c_2 & d_1+d_2 \end{bmatrix} \\\quad \Rightarrow (a_1+a_2) +(b_1+b_2)+ (c_1+c_2) +(d_1+d_2) =(a_1+b_1 +c_1+ d_1) +(a_2+b_2+c_2 +d_2) =0\\ \quad \Rightarrow U_1+U_2 \in U\\ \Rightarrow \bbox[red, 2pt]{\text{Yes, }U \text{ is a subspace.}}$$

解答: $$A =\begin{bmatrix}2&3 \\1& 4 \end{bmatrix}  =\begin{bmatrix}-3&1 \\1& 1 \end{bmatrix}  \begin{bmatrix} 1&0 \\0& 5 \end{bmatrix}  \begin{bmatrix}-1/4& 1/4 \\1/4& 3/4 \end{bmatrix}  \Rightarrow A^{1/2}=\begin{bmatrix}-3&1 \\1& 1 \end{bmatrix}  \begin{bmatrix} \sqrt1&0 \\ 0& \sqrt 5 \end{bmatrix}  \begin{bmatrix}-1/4& 1/4 \\1/4& 3/4 \end{bmatrix} \\=\begin{bmatrix} -3& \sqrt 5 \\1 & \sqrt 5 \end{bmatrix} \begin{bmatrix}-1/4& 1/4 \\1/4& 3/4 \end{bmatrix} = \bbox[red, 2pt] {\begin{bmatrix} \frac{ \sqrt{5} +3}{4} & \frac{3\sqrt{5}-3}{4} \\\frac{\sqrt{5}-1}{4} & \frac{3\sqrt{5}+1}{4} \end{bmatrix}}$$

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解題僅供參考，其他碩士班試題及詳解