國立臺灣師範大學113學年度碩士班招生考試
科目: 工程數學
適用系所: 電機工程學系
解答:$$y=x^m \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-2y= (m^2-m-2)x^m=(m-2)(m+1)x^m =0\\ \Rightarrow m=2,-1 \Rightarrow \bbox[red, 2pt]{y= c_1x^2+{c_2\over x}}$$
解答:$$\cases{P(x,y)=xy\\ Q(x,y)= 2x^2+3y^2-20} \Rightarrow \cases{P_y=x\\ Q_x=4x} \Rightarrow P_y \ne Q_x \Rightarrow \text{ Not Exact} \\ -{P_y-Q_x \over P} ={3\over y} \text{ is dependent on }y \Rightarrow u'={3\over y}u \Rightarrow \text{ integrating factor }u(y)=y^3 \\ \Rightarrow \cases{uP=xy^4\\ uQ= 2x^2y^3 +3y^5-20y^3} \Rightarrow (uP)_y=4xy^3 = (uQ)_x \Rightarrow \text{ Exact} \\ \qquad \Rightarrow \Phi(x,y) = \int xy^4\,dx = \int2x^2y^3+3y^5-20y^3\,dy \\ \qquad \Rightarrow {1\over 2}x^2y^4 + \phi(y)= {1\over 2}x^2y^4+ {1\over 2}y^6-5y^4+ \rho(x)\\ \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{{1\over 2}x^2y^4+ {1\over 2}y^6-5y^4+c_1=0}$$
解答:$$L\left\{ \int_0^t e^\tau \sin(t-\tau)\,d \tau\right\} =L\{e^t\} \cdot L\{ \sin(t)\} ={1\over s-1} \cdot {1\over s^2+1} =\bbox[red, 2pt]{1\over (s-1)(s^2+1)}$$
解答:$$f(x)=\begin{cases}k & 0\lt x\le \pi \\-k & -\pi \le x\lt 0\end{cases},f(x)=f(x+2\pi)\\ \Rightarrow f(-x)=-f(x) \Rightarrow f(x) \text{ is odd} \Rightarrow a_n=0\\ b_n={1\over \pi} \left(\int_{-\pi}^0 -k\sin(nx)\,dx +\int_0^\pi k\sin(nx)\,dx \right) ={k\over \pi}\left( \left. \left[ {1\over n} \cos(nx) \right] \right|_{-\pi}^0 + \left. \left[ -{1\over n}\cos(nx) \right] \right|_{0}^\pi\right) \\ ={k\over \pi}\left({2 \over n}(1-(-1)^n) \right) ={2k\over n\pi }(1-(-1)^n) \Rightarrow \bbox[red, 2pt]{\cases{a_n=0, n=0,1,2,\dots\\ b_n= {2k\over n\pi }(1-(-1)^n),n=1,2,\dots}}$$
解答:$$\textbf{1)}\; \bbox[red, 2pt]{true} \\\textbf{2)}\; \bbox[red, 2pt]{\text{false}}, rref(A)=I \Rightarrow A \text{ is full rank} \\ \textbf{3)}\; \bbox[red, 2pt]{\text{true}}, AA^T=I \Rightarrow \det(A)^2=1 \Rightarrow |\det(A)|=1 \\ \textbf{4)}\; \bbox[red, 2pt]{\text{false}}, A= \begin{bmatrix}2 & 0 \\0 & 2 \end{bmatrix} \Rightarrow \text{rref}(A)=\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix} \Rightarrow |\det(A)| \ne |\det(rref(A))| \\\textbf{5)}\; \bbox[red, 2pt]{\text{true}} \text{ A basis is a linearly independent spanning set.} \\\textbf{6)}\; \bbox[red, 2pt]{\text{false}}, \det(A) =0 \Rightarrow \det(A)\det(A^{-1})=0 \ne \det(I)=1$$
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