國立嘉義大學113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:$$\cases{x_1+x_2+2x_3+6x_4= 11\\ 2x_1+3x_2+6x_3+19x_4=36\\ 3x_2+4x_3+15x_4=28\\ x_1-x_2-x_3-6x_4=-12} \Rightarrow \left[ \begin{matrix} 1 & 1 & 2 & 6 \\2 & 3 & 6 & 19 \\0 & 3 & 4 & 15 \\1 & -1 & -1 & -6 \end{matrix} \right] \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} =\begin{bmatrix}11\\36\\ 28\\-12 \end{bmatrix} \\\Rightarrow \left[ \begin{array}{rrrr| r}1 & 1 & 2 & 6 &11\\2 & 3 & 6 & 19 &36\\0 & 3 & 4 & 15 &28\\1 & -1 & -1 & -6 & -12\end{array} \right] \xrightarrow{R_2-2R_1\to R_2,R_4-R_1\to R_4} \left[ \begin{array}{rrrr| r}1 & 1 & 2 & 6 & 11\\0 & 1 & 2 & 7 & 14\\0 & 3 & 4 & 15 & 28\\0 & -2 & -3 & -12 & -23\end{array} \right] \\ \xrightarrow{R_1-R_2\to R_1,R_3-3R_2\to R_3, R_4+2R_2\to R_4} \left[ \begin{array}{rrrr| r}1 & 0 & 0 & -1 & -3\\0 & 1 & 2 & 7 & 14\\0 & 0 & -2 & -6 & -14\\0 & 0 & 1 & 2 & 5\end{array} \right] \xrightarrow{R_3/(-2)\to R_3} \\ \left[ \begin{array}{rrrr| r}1 & 0 & 0 & -1 & -3\\0 & 1 & 2 & 7 & 14\\0 & 0 & 1 & 3 & 7\\0 & 0 & 1 & 2 & 5\end{array} \right] \xrightarrow{R_2-2R_3\to R_2,R_4-R_3\to R_4} \left[ \begin{array}{rrrr| r}1 & 0 & 0 & -1 & -3\\0 & 1 & 0 & 1 & 0\\0 & 0 & 1 & 3 & 7\\0 & 0 & 0 & -1 & -2\end{array} \right]\\ \xrightarrow{R_1-R_4\to R_1,R_2+R_4\to R_2,R_3+3R_4\to R_3} \left[ \begin{array}{rrrr| r}1 & 0 & 0 & 0 & -1\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 1\\0 & 0 & 0 & -1 & -2\end{array} \right] \Rightarrow \bbox[red, 2pt]{ \cases{x_1=-1\\ x_2=-2\\ x_3=1\\ x_4=2}}$$
解答:$$\textbf{a. } y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow \bbox[red, 2pt]{y_h= c_1\cos x+c_2\sin x}\\ \textbf{b. }y''+y=2e^{ix} = 2\cos x+ 2i\sin x \Rightarrow y_p=A x\cos x+B x\sin x \\ \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+Bx\cos x\\ \Rightarrow y_p''=-2A\sin x -Ax\cos x+2B\cos x -Bx\sin x \\ \Rightarrow y_p''+y_p= -2A\sin x+2B\cos x= 2\cos x+2i\sin x \Rightarrow \cases{A=-i \\ B=1 } \\ \Rightarrow y_p=-ix \cos x+ x\sin x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1\cos x+c_2\sin x-ix \cos x+ x\sin x}$$
解答:$$\textbf{a. } y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2}\\ \Rightarrow x^2y''-xy'+y =m(m-1)x^m-mx^m+x^m=(m^2-2m+1)x^m=0\\ \Rightarrow m=1 \Rightarrow \bbox[red, 2pt]{y_h=c_1x+c_2 x\ln x} \\\textbf{b. } \cases{y_1=x\\y_2=x\ln x\\ r(x)=\ln x/x^2} \Rightarrow W(x)=\begin{vmatrix}y_1& y_2\\ y_1'& y_2' \end{vmatrix} =x\\ \text{Applying variations of parameters, }y_p =-x \int{x\ln x\cdot {\ln x\over x^2} \over x}\,dx + x\ln x\int {x\cdot {\ln x\over x^2} \over x} \,dx \\=-x\int{(\ln x)^2\over x^2}\,dx+ x\ln x\int{\ln x\over x^2}\,dx =\ln x+2 \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1x+c_2x\ln x+\ln x+2}$$
解答:$$\vec v_1=\begin{pmatrix}1\\ 1\\1 \end{pmatrix}, \vec v_2=\begin{pmatrix}2\\ 0\\1 \end{pmatrix}, \vec v_3= \begin{pmatrix}2\\ 4\\5 \end{pmatrix}\\ \vec u_1=\vec v_1 \Rightarrow \vec e_1={\vec u_1\over |\vec u_1|} ={\sqrt 3\over 3}\begin{pmatrix}1\\ 1\\1 \end{pmatrix} \\ \vec u_2=\vec v_2-(\vec e_1\cdot \vec v_2)\vec e_1 =\begin{pmatrix}1\\ -1\\0 \end{pmatrix} \Rightarrow \vec e_2={\vec u_2\over |\vec u_2|} ={\sqrt 2\over 2}\begin{pmatrix}1\\ -1\\0 \end{pmatrix}\\ \vec u_3=\vec v_3-(\vec v_3\cdot \vec e_1)\vec e_1 -(\vec v_3\cdot \vec e_2)\vec e_2={2\over 3} \begin{pmatrix} -1\\ -1\\2 \end{pmatrix} \Rightarrow \vec e_3={\vec u_3\over |\vec u_3|}= {\sqrt 6 \over 6}\begin{pmatrix}-1\\ -1\\2 \end{pmatrix} \\ \Rightarrow \text{orthonormal basis} = \bbox[red, 2pt]{\left\{\begin{pmatrix} \sqrt 3/3\\ \sqrt 3/3\\\sqrt 3/3 \end{pmatrix}, \begin{pmatrix} \sqrt 2/2\\ -\sqrt 2/2\\0 \end{pmatrix}, \begin{pmatrix}-\sqrt 6/6\\ -\sqrt 6/6\\\sqrt 6/3 \end{pmatrix} \right\}}$$
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解題僅供參考，碩士班歷年試題及詳解
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