國立雲林科技大學110學年度碩士班招生考試
系所:電機系
科目:工程數學(2)
解答:$$\textbf{(1)}\; y'=10 \sin(2x) \Rightarrow y=\int 10\sin(2x)\,dx = -5\cos(2x)+c_1 \Rightarrow \bbox[red, 2pt]{y=-5\cos(2x)+c_1} \\\textbf{(2)}\; \text{Integration factor }I(x)=e^{\int 3x^2\,dx} =e^{x^3} \Rightarrow I(x)y'+3x^2yI(x)=2e^{-x^3}I(x) \\\quad \Rightarrow e^{x^3}y' +3x^2 e^{x^3}y=2 \Rightarrow (e^{x^3}y)'=2 \Rightarrow e^{x^3}y=2x +c_1 \Rightarrow \bbox[red, 2pt]{y=2xe^{-x^3}+c_1e^{-x^3}} \\\textbf{(3)}\; \text{Let }v(x)={1\over y} \Rightarrow v'=-{y'\over y^2} \Rightarrow y'=-y^2v'=-{v'\over v^2} \Rightarrow -{v'\over v^2}-{1\over v}={e^x\over v^2} \\\quad \Rightarrow e^xv'+e^x v=-e^{2x} \Rightarrow (e^x v)'=-e^{2x} \Rightarrow e^xv = -{1\over 2}e^{2x} +c_1 \Rightarrow v=-{1\over 2}e^x +c_1e^{-x} \\\quad \Rightarrow y={1\over -{1\over 2}e^x +c_1e^{-x}} \Rightarrow \bbox[red, 2pt]{y={-2e^x \over e^{2x}+c_2}}$$
解答:$$\textbf{(1)}\; F(s)=L\left\{ t^2-e^{-2t}+\cos(2t) \right\}=\bbox[red, 2pt]{{2 \over s^3}-{1\over s+2}+{s\over s^2+2^2}} \\ \textbf{(2)}\; L^{-1}\left\{ {12s \over (s-1)(s+1)(s+2)}\right\} =L^{-1}\left\{ { 2\over s-1} +{6\over s+1} -{8\over s+2}\right\} = \bbox[red, 2pt]{2e^t+ 6e^{-t}-8e^{-2t}} \\ \textbf{(3)}\; F(s)={2\over s(1+e^{-as})} ={2\over s}\cdot {1\over 1+e^{-as}} \\ {1\over 1+e^{-as}} ={1\over 1-(-e^{-as})} =1-e^{-as}+e^{-2as}-e^{-3as}+\cdots =\sum_{n=0}^\infty (-1)^ne^{-nas }\\ \text{If }L\{ f(t)\} =F(s) \Rightarrow {1\over 1+e^{-as}} F(s)= \sum_{n=0}^\infty (-1)^ne^{-nas } F(s) \\ \Rightarrow L^{-1}\left\{{1\over 1+e^{-as}} F(s) \right\} =L^{-1}\left\{\sum_{n=0}^\infty (-1)^ne^{-nas } F(s) \right\} = \sum_{n=0}^\infty (-1)^n f(t-na) u(t-na)\\ \text{Now, }L^{-1}\left\{ {2\over s(1+e^{-as})}\right\} = L^{-1}\left\{ {1\over 1+e^{-as}}\cdot {2\over s}\right\} =\bbox[red, 2pt]{2\sum_{n=0}^\infty (-1)^n u(t-na)}$$
解答:$$\cases{2x_1+ 4x_2+6x_3=18\\ 4x_1+5x_2 +6x_3=24\\ 2x_1+7x_2+12x_3=40} \Rightarrow \left[ \begin{array}{rrr|r}2 & 4 & 6 & 18\\4 & 5 & 6 & 24\\2 & 7 & 12 & 40\end{array} \right] \xrightarrow{R_2-2R_1 \to R_2, R_3-R_1 \to R_3} \\ \left[ \begin{array}{rrr|r} 2 & 4 & 6 & 18\\0 & -3 & -6 & -12\\0 & 3 & 6 & 22\end{array} \right] \xrightarrow{-(1/3)R_2 \to R_2} \left[ \begin{array}{rrr|r} 2 & 4 & 6 & 18\\0 & 1 & 2 & 4\\0 & 3 & 6 & 22 \end{array} \right] \xrightarrow{R_3-3R_2 \to R_3, R_1-4R_2\to R_1} \\ \left[ \begin{array}{rrr|r}2 & 0 & -2 & 2\\0 & 1 & 2 & 4\\0 & 0 & 0 & 10 \end{array} \right] \xrightarrow{(1/2)R_1\to R_1, (1/10)R_3\to R_3} \left[ \begin{array}{rrr|r}1 & 0 & -1 & 1\\0 & 1 & 2 & 4\\0 & 0 & 0 & 1\end{array} \right] \xrightarrow{R_1-R_3\to R_1,R_2-4R_3\to R_2} \\ \left[ \begin{array}{c}1 & 0 & -1 & 0\\0 & 1 & 2 & 0\\0 & 0 & 0 & 1\end{array} \right] \Rightarrow \text{rwo-reduce form }\bbox[red, 2pt]{\left[ \begin{array}{c}1 & 0 & -1 & 0\\0 & 1 & 2 & 0\\0 & 0 & 0 & 1\end{array} \right]} \Rightarrow \cases{x_1-x_3=0\\ x_2+2x_3=0\\ 0=1} \Rightarrow \bbox[red, 2pt]{No}\text{ solution}$$
解答:$$A=\left[\begin{matrix}1 & 2 & -2 & 1\\3 & 6 & -5 & 4\\1 & 2 & 0 & 3\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 2 & 0 & 3\\0 & 0 & 1 & 1\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \cases{x_1+2x_2+3x_4=0\\ x_3+x_4=0} \\ \Rightarrow \mathbf x=x_2\begin{pmatrix}-2 \\1\\0\\0 \end{pmatrix} +x_4\begin{pmatrix}-3 \\0\\-1\\1 \end{pmatrix} \Rightarrow \bbox[red, 2pt]{N(A)=\left\{s\begin{pmatrix}-2 \\1\\0\\0 \end{pmatrix} +t\begin{pmatrix}-3 \\0\\-1\\1 \end{pmatrix} \mid s,t\in \mathbb R\right\}}, \\\bbox[red, 2pt]{rank(A)=2, nullity(A)=2}$$
解答:$$\textbf{(a)}\; (0,0,0)\not \in W_1 \Rightarrow W_1 \text{is }\bbox[red, 2pt]{\text{NOT}}\text{ a subspace of }\mathbb R^3 \\ \textbf{(b)}\; \cases{u=(u_1,u_1+u_3,u_3)\in W_2\\ v=(v_1,v_1+v_3, v_3) \in W_2} \Rightarrow \cases{u+v=(u_1+v_1,u_1+v_1 +u_3+v_3, u_3+v_3) \\ cu=(cu_1,c(u_1+u_3), cu_3)}\\ \quad \Rightarrow \cases{u+v\in W_2\\ cu\in W_2} \Rightarrow W_2 \bbox[red, 2pt]{\text{IS}}\text{ a subspace of }\mathbb R^3$$
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解題僅供參考,其他歷年試題及詳解
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