110年雲科大電機碩士班-工程數學詳解

 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： $$y''-2y'=0 \Rightarrow \lambda^2-2\lambda=0 \Rightarrow \lambda=2,0 \Rightarrow y_h=c_1e^{2x}+c_2\\ \text{Applying variation of parameters, let} \cases{y_1=e^{2x}\\ y_2=1} \Rightarrow W=\begin{vmatrix}y_1 & y_2 \\y_1' & y_2' \end{vmatrix}  = \begin{vmatrix}e^{2x} & 1 \\2e^{2x} & 0 \end{vmatrix} = -2e^{2x} \\ \Rightarrow y_p=-e^{2x} \int{ -6+e^x\over -2e^{2x}}\,dx +\int{ e^{2x}(-6+e^x)\over -2e^{2x}}\,dx ={3\over 2}+3x-e^x \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{2x}+c_3+3x-e^x}$$

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： $$L\{ y''+2y'+5y\} =6L\{\delta(t-2)\} \Rightarrow s^2Y(s)+2sY(s)+5Y(s)=6e^{-2s} \\ \Rightarrow Y(s)={6\over s^2+2s+5}e^{-2s} ={2\cdot 3\over (s+1)^2+2^2} e^{-2s} \\\Rightarrow y(t)=L^{-1}\{ Y(s) \}\Rightarrow \bbox[red, 2pt]{ y(t) =u(t-2)\cdot 3e^{-(t-2)} \sin(2(t-2))}$$

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： $$\cases{A=\left[ \begin{matrix}1 & 1\\1 & 2\\1 & 3\end{matrix} \right] \\[1ex] b=\left[ \begin{matrix} 0\\1\\3 \end{matrix} \right]} \Rightarrow \cases{A^TA=\left[\begin{matrix}3 & 6\\6 & 14\end{matrix}\right] \\[1ex] A^Tb=\begin{bmatrix}4 \\11 \end{bmatrix}} \Rightarrow \mathbf x'=(A^TA)^{-1}(A^Tb) =\left[\begin{matrix}\frac{7}{3} & -1\\-1 & \frac{1}{2}\end{matrix}\right] \begin{bmatrix}4 \\11 \end{bmatrix} \\ \Rightarrow \mathbf x'=\bbox[red, 2pt]{\left[\begin{matrix}- \frac{5}{3}\\\frac{3}{2}\end{matrix}\right]}$$

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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