114年中正大學機械碩士班-工程數學詳解

國立中正大學114學年度碩士班招生考試試題

科目名稱:工程數學
系所組別:機械工程學系-乙組

解答: $$y'=2y^2e^x \Rightarrow {1\over 2y^2}dy =e^xdx \Rightarrow \int {1\over 2y^2}dy = \int e^x\,dx \Rightarrow -{1\over 2y} =e^x+c_1 \\ \Rightarrow 2y={-1\over e^x+c_1} \Rightarrow \bbox[red, 2pt]{y={-1\over 2e^x+2c_1}}$$

解答: $$L\{y'\}-3L\{y\}=L\{1\} \Rightarrow sY(s)-y(0)-3Y(s)={1\over s} \Rightarrow (s-3)Y(s)={1\over s}+1 \\ \Rightarrow \bbox[red, 2pt]{Y(s) ={1\over s(s-3)} +{1\over s-3}}$$

解答: $$\textbf{(1) } \mathcal F\{f(x)\} = \int_{-\infty}^\infty f(x)e^{-j\omega x}\,dx = \int_{-a}^a e^x\cdot e^{-j\omega x}\,dx= \int_{-a}^a   e^{(1-j\omega) x}\,dx =\left. \left[ {1\over 1-j\omega} e^{(1-j\omega) x}\right] \right|_{-a}^a \\\qquad = \bbox[red, 2pt]{{1\over 1-j\omega} \left( e^{(1-j\omega)a} -e^{(j\omega -1)a}\right)} \\\textbf{(2) } \mathcal F\{f(x)\} = \int_0^a xe^{-j\omega x}\,dx =\left. \left[ {1+j\omega x\over \omega^2}e^{-j\omega x} \right] \right|_0^a = \bbox[red, 2pt]{{1+j\omega a\over \omega^2}e^{-j\omega a}-{1\over \omega^2}} \\\textbf{(3) } \mathcal F\{f(x)\} = \int_{-1}^1 |x|e^{-j\omega x}\,dx = 2\int_{0}^1 xe^{-j\omega x}\,dx = \bbox[red, 2pt]{2\left( {1+j\omega \over \omega^2} e^{-j\omega} -{1\over \omega^2}\right) }$$

解答: $$\textbf{(1) } \det(A) =\begin{vmatrix} 2&{1\over 10}& {33\over 10} \\ {8\over 5} & {22\over 5} & {1\over 2} \\ {3\over 10}& -{43\over 10}& {14\over 5}\end{vmatrix} ={1447\over 1000} \Rightarrow \cases{C_{1,1} =\begin{vmatrix} {22\over 5} & {1\over 2} \\ -{43\over 10}& {14\over 5}\end{vmatrix} ={1447\over 100} \\C_{1,2} =-\begin{vmatrix} {8\over 5} & {1\over 2} \\ {3\over 10}& {14\over 5}\end{vmatrix} =-{433\over 100} \\ C_{1,3} =\begin{vmatrix} {8\over 5} & {22\over 5} \\ {3\over 10}& -{43\over 10} \end{vmatrix} =-{41\over 5} \\ C_{2,1} = -\begin{vmatrix} {1\over 10}& {33\over 10} \\ -{43\over 10}& {14\over 5}\end{vmatrix} =-{1447\over 100} \\ C_{2,2} =\begin{vmatrix} 2& {33\over 10} \\ {3\over 10} & {14\over 5}\end{vmatrix} ={461\over 100} \\C_{2,3} = -\begin{vmatrix} 2&{1\over 10} \\ {3\over 10}& -{43\over 10} \end{vmatrix} ={863\over 100} \\ C_{3,1} =\begin{vmatrix} {1\over 10}& {33\over 10} \\ {22\over 5} & {1\over 2} \end{vmatrix} =-{1447\over 100} \\ C_{3,2} =-\begin{vmatrix} 2& {33\over 10} \\ {8\over 5} & {1\over 2} \end{vmatrix} ={107\over 25} \\C_{3,3} =\begin{vmatrix} 2&{1\over 10} \\ {8\over 5} & {22\over 5} \end{vmatrix} ={216\over 25}} \\\qquad \Rightarrow A^{-1} ={1\over \det(A)} \begin{bmatrix} C_{1,1}& C_{1,2}& C_{1,2} \\ C_{2,1} & C_{2,2}& C_{2,3} \\ C_{3,1}& C_{3,2} &C_{3,3}\end{bmatrix} \Rightarrow \bbox[red, 2pt]{A^{-1}=\left[ \begin{matrix} 10 & -10 & -10 \\\frac{-4330}{1447} & \frac{4610}{1447} & \frac{4280}{1447} \\\frac{-8200}{1447} & \frac{8630}{1447} & \frac{8640}{1447} \end{matrix} \right]}\\ \qquad \det(A-\lambda I) =-\lambda^3+{46\over 5}\lambda^2-{693\over 25}\lambda+{1447\over 1000}=0 \Rightarrow  \bbox[red, 2pt]{\lambda \approx 0.053,4.573\pm 2.514i} \\\textbf{(2) }\det(B) =10375 \Rightarrow \cases{C_{1,1}= 2925\\ C_{1,2}=-1650\\ C_{1,3}=-50\\ C_{2,1} =-1650 \\ C_{2,2} =1250\\ C_{2,3} =-25\\ C_{3,1} =-50\\ C_{3,2} =-25\\ C_{3,3} =125} \Rightarrow B^{-1} ={1\over 10375} \left[ \begin{matrix}2925 & -1650 & -50 \\-1650 & 1250 & -25 \\-50 & -25 & 125 \end{matrix} \right] \\ \qquad \Rightarrow \bbox[red, 2pt]{B^{-1} =\left[ \begin{matrix} \frac{117}{415} & \frac{-66}{415} & \frac{-2}{415} \\\frac{-66}{415} & \frac{10}{83} & \frac{-1}{415} \\ \frac{-2}{415} & \frac{-1}{415} & \frac{1}{83} \end{matrix} \right]}\\ \qquad \det(B-\lambda I) =-\lambda^3 +140\lambda^2-4300\lambda+10375 =0 \Rightarrow \bbox[red, 2pt]{\lambda \approx 2.6,40.8,96.6}$$

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解題僅供參考，其他碩士班試題及詳解