114年嘉義大學電機碩士班-工程數學詳解

 國立嘉義大學 114 學年度電機工程學系碩士班招生考試

科目: 工程數學 (每題 25 分，共 100 分)

解答:$$W=\{(a,-2a, b) \mid a,b \in \mathbb R\} = span\{(1,-2,0), (0,0,1)\} \Rightarrow W^\bot = span\{(2,1,0)\} \\ \Rightarrow \text{proj}_{W^\bot}(X)= {X\cdot u\over ||u||^2}u ={(2,4,-1)\cdot (2,1,0)\over 2^2+1^2+0^2}(2,1,0)= \left( {16\over 5},{8\over 5},0\right) \\ \Rightarrow \text{distance }=||\text{proj}_{W^\bot}(X)|| =\sqrt{{256\over 25}+ {64\over 25}} = \bbox[red, 2pt]{8\sqrt 5\over 5}$$

解答:$$\cases{x_1-2x_2+3x_3+x_4=1\\ 2x_1-3x_2+2x_3 -x_4=4\\ 3x_1-5x_2+ 5x_3=5\\ x_1-x_2-x_3-2x_4=3} \Rightarrow \begin{bmatrix} 1&-2& 3& 1\\ 2&-3& 2& -1\\ 3& -5& 5& 0\\ 1& -1&-1 &-2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}= \begin{bmatrix} 1\\4\\5\\3\end{bmatrix} \\ A=\left[ \begin{array}{rrrr|r} 1&-2& 3& 1&1\\ 2&-3& 2& -1&4\\ 3& -5& 5& 0& 5\\ 1& -1&-1 &-2& 3\end{array} \right] \Rightarrow rref(A) =\left[ \begin{array}{rrrr|r} 1&0& -5& -5& 5\\ 0&1& -4& -3& 2\\ 0& 0& 0& 0& 0\\  0& 0& 0& 0& 0\end{array} \right] \Rightarrow \cases{x_1-5x_3-5x_4=5\\ x_2-4x_3-3x_4=2} \\ \Rightarrow \text{ solution set: } \bbox[red, 2pt]{\{(5s+5t+5,4s+3t+2,s,t) \mid s,t\in \mathbb R\} }\\ \Rightarrow \text{kernel space: }\{(5s+5t,4s+3t,s,t) \mid s,t\in \mathbb R\} = \bbox[red, 2pt]{span\left\{ \begin{pmatrix}5\\4\\1\\0 \end{pmatrix},\begin{pmatrix}5\\3\\0\\1 \end{pmatrix} \right\}} \\ \Rightarrow \text{Geometrically,the kernel space is a plane passing through the origin.} \\ \text{Geometrically, the solution set is a plane and parallel to the kernel space.}\\\text{ It is a direct translation of the kernel plane, shifted away from the origin by }\begin{pmatrix}5\\2\\0\\0 \end{pmatrix}.$$

解答:$$\textbf{(a) } \bbox[red, 2pt]{\text{Yes, }F \text{ is a linear ODE}} \\\textbf{(b) } y'\sin x+(\cos x)y=1 \Rightarrow (y\sin x)'=1 \Rightarrow y\sin x = \int 1\,dx = x+c_1 \Rightarrow \bbox[red, 2pt]{y={x+c_1\over \sin x}}$$

解答:$$\textbf{a. } \lambda^2-2\lambda+1=0 \Rightarrow (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow \bbox[red, 2pt]{y=c_1e^x +c_2xe^x} \\\textbf{b. } \cases{y_1=e^x\\ y_2 =xe^x \\R(x)=6xe^x} \Rightarrow W= \begin{vmatrix} y_1 &y_2\\ y_1'& y_2'\end{vmatrix} =\begin{vmatrix} e^x & xe^x\\ e^x &  e^x+xe^x\end{vmatrix} =e^{2x} \\\quad \Rightarrow y_p =-y_1\int{ y_2R\over W}\,dx+ y_2\int{y_1R\over W}\,dx =-e^x\int 6x^2\,dx +xe^x \int6x\,dx =-2x^3e^x+3x^3e^{x}=x^3e^x \\\quad \Rightarrow \bbox[red, 2pt] {y=c_1e^x+ c_2xe^x+x^3e^x} \\ \textbf{c. } \cases{y_1=e^x\\ y_2 =xe^x \\R(x)=6e^x \cos x} \Rightarrow W=e^{2x} \Rightarrow y_p=-e^x \int 6x\cos x\,dx +xe^{x} \int 6\cos x\,dx \\\quad =-e^x(6x\sin x+6\cos x)+ xe^x(6\sin x) = -6e^x \cos x \Rightarrow \bbox[red, 2pt]{y=c_1e^x +c_2xe^x-6e^x \cos x}$$

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