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:$$\textbf{a. } y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow \bbox[red, 2pt]{y_h= c_1\cos x+c_2\sin x}\\ \textbf{b. }y''+y=2e^{ix} = 2\cos x+ 2i\sin x \Rightarrow y_p=A x\cos x+B x\sin x \\ \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+Bx\cos x\\ \Rightarrow y_p''=-2A\sin x -Ax\cos x+2B\cos x -Bx\sin x \\ \Rightarrow y_p''+y_p= -2A\sin x+2B\cos x= 2\cos x+2i\sin x \Rightarrow \cases{A=-i \\ B=1 } \\ \Rightarrow y_p=-ix \cos x+ x\sin x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1\cos x+c_2\sin x-ix \cos x+ x\sin x}$$
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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