國立中正大學111學年度碩士班招生考試
科目:工程數學
系所組別: 機械工程學系乙組
解答: $$\textbf{(a)}\; \frac{d ^3 y}{dx^3} -y=0 \Rightarrow \lambda^3-1=0 \Rightarrow (\lambda-1)(\lambda^2+ \lambda+1)=0 \Rightarrow \lambda=1, {-1\pm \sqrt 3i\over 2} \\ \quad \Rightarrow \bbox[red, 2pt]{y_h=c_1e^x +e^{-x/2}(c_2\cos {\sqrt 3\over 2}x + c_3 \sin {\sqrt 3\over 2}x)} \\ \textbf{(b)}\; y_p=Ae^{-x} \Rightarrow y_p'=-Ae^{-x } \Rightarrow y_p''=Ae^{-x} \Rightarrow y_p''' =-Ae^{-x} \Rightarrow y_p'''-y_p=-2Ae^{-x} =3e^{-x} \\ \quad \Rightarrow A= -{3\over 2} \Rightarrow y_p=-{3\over 2}e^{-x} \Rightarrow y=y_h+y_p \\\quad \Rightarrow y=c_1e^x +e^{-x/2}(c_2\cos {\sqrt 3\over 2}x + c_3 \sin {\sqrt 3\over 2}x)-{3\over 2}e^{-x} \\\quad \Rightarrow y'= c_1e^x +e^{-x/2} \left[\left( -{1 \over 2} c_2 +{\sqrt 3\over 2}c_3\right) \cos{\sqrt 3\over 2}x +\left( -{\sqrt 3\over 2}c_2-{1\over 2} c_2\right)\sin{\sqrt 3\over 2}x \right]+{3\over 2}e^{-x} \\ \Rightarrow y''=c_1e^x +e^{-x/2} \left[ \left( -{1\over 2}c_2-{\sqrt 3\over 2}c_3\right) \cos{\sqrt 3\over 2}x +\left( {\sqrt 3\over 2}c_2 -{1\over 2}c_3\right) \sin {\sqrt 3\over 2}x\right]-{3\over 2}e^{-x} \\ \text{initial conditions:}\cases{y(0)=0\\ y'(0)=1\\ y''(0)=0} \Rightarrow \cases{c_1+c_2-3/2=0\\ c_1-c_2/2+ \sqrt 3c_3/2+3/2=1\\ c_1-c_2/2-\sqrt 3c_3/2-3/2=0} \Rightarrow \cases{c_1=5/6\\ c_2=2/3\\ c_3=-2\sqrt 3/3} \\ \Rightarrow \bbox[red, 2pt]{y={5\over 6} e^x+e^{-x/2} \left( {2\over 3}\cos {\sqrt 3\over 2}x-{2\sqrt 3\over 3}\sin {\sqrt 3\over 2}x \right)-{3\over 2}e^{-x}}$$
解答: $$\textbf{(a)}\; L\{y''\}-L\{y\}=L\{e^{-t}\} \Rightarrow s^2Y(s)-3s-Y(s)={1\over s+1}\\\quad \Rightarrow Y(s)={1\over (s+1)(s^2-1)}+{3s\over s^2-1} ={3s^2+3s+1\over (s+1)(s^2-1)}\\={7\over 4(s-1)} +{5\over 4(s+1)}-{1 \over 2(s+1)^2} \\ \Rightarrow y(t)=L^{-1}\{Y(s)\} ={7 \over 4}L^{-1}\{ {1\over s-1}\}+{5\over 4}L^{-1}\{{1\over s+1 }\}- {1\over 2}L^{-1}\{{1\over (s+1)^2}\} \\\quad \Rightarrow \bbox[red, 2pt]{y(t)={7\over 4}e^t +{5\over 4}e^{-t} -{1 \over 2}te^{-t}} \\\textbf{(b)}\; L\{y'\}+2L\{y\}=L\{2(u(t)-u(t-2))\} \Rightarrow sY(s)+2Y(s)={2\over s}(1-e^{-2s}) \\\quad \Rightarrow Y(s)={2\over s^2+2s}-{2e^{-2s}\over s^2+2s} ={1\over s}-{1\over s+2}-\left({1\over s}-{1\over s+2} \right)e^{-2s} \\ \Rightarrow y(t)= L^{-1}\{Y(s)\} \Rightarrow \bbox[red, 2pt]{y(t) =u(t)-e^{-2t}-u(t-2)(1-e^{-2(t-2)})}$$
解答: $$\textbf{(a)}\; f(s)= \begin{vmatrix} 2-s& -1 & -1 &0\\ -1& 3-s& -1 & -1\\ -1 & -1 & 3-s& -1\\ 0 & -1& -1 & 2-s\end{vmatrix} \xrightarrow{R_1-R_4\to R_1,R_2-R_4\to R_2,R_3-R_4 \to R_3} \begin{vmatrix} 2-s& 0 & 0 &s-2\\ -1& 4-s& 0 & s-3\\ -1 & 0 & 4-s& s-3\\ 0 & -1& -1 & 2-s\end{vmatrix} \\ \quad \Rightarrow f(s)=(2-s)\begin{vmatrix} 4-s& 0 & s-3\\ 0 & 4-s& s-3\\ -1& -1 & 2-s\end{vmatrix}-(s-2) \begin{vmatrix} -1& 4-s& 0 \\ -1 & 0 & 4-s \\ 0 & -1& -1 \end{vmatrix} \\\quad =(s-4)^2(s-2)^2+ 2(s-2)(s-3)(s-4) -2(s-2)(s-4) =s^4-10s^3+32s^2-32s, \bbox[red, 2pt]{QED} \\ \textbf{(b)}\; f(s)=(s-4)^2(s-2)^2+ 2(s-2)(s-3)(s-4)-2(s-2)(s-4) \\\quad = (s-2)(s-4)[(s-2)(s-4)+ 2(s-3)-2] =(s-2)(s-4)(s^2-4s)\\ \quad =s(s-2)(s-4)^2 \Rightarrow \text{If }f(s)=0, \text{ then }s=0,2,4 \Rightarrow \text{eigenvalues of }M:\bbox[red, 2pt]{0,2,4} \\ \textbf{(c)}\; \det(M)=0 \Rightarrow M \text{ is }\bbox[red, 2pt]{ \text{NOT invertible}} \\\textbf{(d)}\;(M-4I) v=0 \Rightarrow \begin{bmatrix}-2 & -1 & -1 & 0 \\-1 & -1 & -1 & -1 \\-1 & -1 & -1 & -1 \\0 & -1 & -1 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3 \\x_4 \end{bmatrix} =0 \Rightarrow \cases{x_1-x_4=0 \\ x_2+x_3+2x_4=0} \\ \quad \Rightarrow v=x_3\begin{bmatrix}0 \\-1 \\ 1\\0\end{bmatrix} +x_4 \begin{bmatrix}1 \\-2 \\ 0 \\1 \end{bmatrix} \Rightarrow \text{eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix}0 \\-1 \\ 1\\0\end{bmatrix},\begin{bmatrix}1 \\-2 \\ 0 \\1 \end{bmatrix} } \\\textbf{(e)}\; M= \begin{bmatrix} 1 & -1 & 0 & 1 \\1 & 0 & -1 & -2 \\1 & 0 & 1 & 0 \\1 & 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}0 & 0 & 0 & 0 \\0 & 2 & 0 & 0 \\0 & 0 & 4 & 0 \\0 & 0 & 0 & 4 \end{bmatrix} \begin{bmatrix}\frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\\frac{-1}{2} & 0 & 0 & \frac{1}{2} \\\frac{-1}{4} & \frac{-1}{4} & \frac{3}{4} & \frac{-1}{4} \\\frac{1}{4} & \frac{-1}{4} & \frac{-1}{4} & \frac{1}{4} \end{bmatrix} \\ \quad \Rightarrow \bbox[red, 2pt]{\text{Yes, M can be diagonalized.}}$$
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