115年中正大學機械碩士班-工程數學詳解

國立中正大學115學年度碩士班招生考試

科目名稱:工程數學
系所組別:機械工程學系-乙組

解答:$$y''+y'-2y=0 \Rightarrow r^2+r-2=0 \Rightarrow (r+2)(r-1)=0 \Rightarrow r=-2,1\\ \Rightarrow y=c_1e^{-2x}+c_2 e^x \Rightarrow y'=-2c_1e^{-2x}+c_2e^x \Rightarrow \cases{y(0)=c_1+c_2= 4\\ y'(0)=-2c_1+c_2=-5} \Rightarrow \cases{c_1=3\\ c_2=1} \\ \Rightarrow \bbox[red, 2pt]{y=3e^{-2x}+e^x}$$

解答:$$L\{y''\} +L\{y'\}+L\{9y\} =0 \Rightarrow  [s^2Y(s) - sy(0) - y'(0)] + [sY(s) - y(0)] + 9Y(s) = 0  \\ \Rightarrow  [s^2Y(s) - 0.16s - 0] + [sY(s) - 0.16] + 9Y(s) = 0  \Rightarrow  Y(s) = \frac{0.16(s + 1)}{s^2 + s + 9} \\=   \frac{0.16(s + 0.5)}{(s + 0.5)^2 + 8.75} + \frac{0.08}{(s + 0.5)^2 + 8.75}  \\ = 0.16 \cdot \frac{s + 0.5}{(s + 0.5)^2 + \left(\frac{\sqrt{35}}{2}\right)^2} + \frac{0.16}{\sqrt{35}} \cdot \frac{\frac{\sqrt{35}}{2}}{(s + 0.5)^2 + \left(\frac{\sqrt{35}}{2}\right)^2}  \\ \Rightarrow y(t)=L^{-1}\{Y(s)\} \Rightarrow \bbox[red, 2pt]{y(t) = 0.16 e^{-0.5t} \cos\left(\frac{\sqrt{35}}{2}t\right) + \frac{0.16}{\sqrt{35}} e^{-0.5t} \sin\left(\frac{\sqrt{35}}{2}t\right)} $$

解答:$$\textbf{(a) } Q(x_1, x_2, x_3) = -4x_1^2 - 7x_2^2 - 4x_3^2 + 2x_1x_2 + 6x_2x_3 = \begin{bmatrix}x_1& x_2& x_3 \end{bmatrix} \begin{bmatrix}a_{11} & a_{12} & a_{13} \\a_{12} & a_{22} & a_{23} \\a_{13} & a_{23} & a_{33}\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} \\ \Rightarrow \cases{a_{11}=-4\\a_{22}=-7\\ a_{33}=-4\\ 2a_{12}=2 \Rightarrow a_{12}=1\\ a_{13} =0\\ 2a_{23}= 6 \Rightarrow a_{23}=3} \Rightarrow \bbox[red, 2pt]{Q= \begin{bmatrix}x_1& x_2& x_3 \end{bmatrix} \begin{bmatrix} -4 & 1 & 0 \\ 1 & -7 & 3 \\ 0 & 3 & -4\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} } \\ \textbf{(b) }A= \begin{bmatrix} -4 & 1 & 0 \\ 1 & -7 & 3 \\ 0 & 3 & -4\end{bmatrix}  \Rightarrow \det(A-\lambda I) = -(\lambda+4)(\lambda+9)(\lambda+2)=0 \Rightarrow \cases{\lambda_1=-4 \\ \lambda_2=-9 \\ \lambda_3=-2} \\ \quad \Rightarrow \text{ All eigenvalues are negative} \Rightarrow \bbox[red, 2pt]{\text{negative definite}}$$

解答:$$\textbf{(a) }   F(j\omega) = \int_{-\infty}^{\infty} f(x) e^{-j\omega x} dx =\int_{0}^{1} (1) e^{-j\omega x} dx + \int_{1}^{2} (-x + 2) e^{-j\omega x} dx  \\ \qquad =     \frac{1 - e^{-j\omega}}{j\omega}  +\frac{e^{-j\omega}}{j\omega} - \frac{e^{-j2\omega} - e^{-j\omega}}{\omega^2}   = \bbox[red, 2pt]{\frac{1}{j\omega} + \frac{e^{-j\omega} - e^{-j2\omega}}{\omega^2} } \\ \textbf{(b) } f_1(x): \text{period of }T=2, \text{ the series contain both sine and cosine terms} \\f_2(x): \text{period of }T=1, \text{ the series contain both sine and cosine terms.} \\f_3(x): \text{period of }T=4, \text{ the series contain only cosine terms and a constant term, }\\ \qquad \text{ with no sine terms.}$$

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解題僅供參考，碩士班歷年試題及詳解