113年台北科大土木碩士班乙組-工程數學詳解

 國立台北科技大學113學年度碩士班招生考試

系所組別: 土木工程系土木與防災碩士班乙組
第二節 工程數學

解答: $$\textbf{1.}\; y'+{y\over x}=x^2 \Rightarrow xy'+y=x^3 \Rightarrow (xy)'=x^3 \Rightarrow xy={1\over 4}x^4+c_1 \Rightarrow \bbox[red, 2pt]{y={1\over 4}x^3+{c_1\over x}} \\ \textbf{2.}\; y'''+y'=0 \Rightarrow \lambda^3+\lambda=0 \Rightarrow \lambda(\lambda^2+1)=0 \Rightarrow \lambda=0,\pm i \\ \quad \Rightarrow \bbox[red, 2pt]{y=c_1+c_2\cos x+ c_3\sin x} \\   \textbf{3.}\; y''-3y'=x \Rightarrow e^{-3x}y''-3e^{-3x}y'=xe^{-3x} \Rightarrow (e^{-3x}y')'=xe^{-3x}\\ \quad \Rightarrow e^{-3x}y'=-{1\over 3}xe^{-3x}-{1\over 9}e^{-3x} +c_1 \Rightarrow y'=-{1\over 3}x-{1\over 9}+c_1e^{3x} \\\quad \Rightarrow y=-{1\over 6}x^2-{1\over 9}x+{c_1\over 3}e^{3x}+c_2 \Rightarrow \bbox[red, 2pt]{y=-{1\over 6}x^2-{1\over 9}x+c_3e^{3x}+ c_2} \\  \textbf{4.}\; \lambda^4-4\lambda^3+5\lambda^2-4\lambda+4 =0 \Rightarrow (\lambda-2)^2 (\lambda^2+1)=0 \Rightarrow \lambda=2,\pm i \\\quad \Rightarrow \bbox[red, 2pt]{y=c_1e^{2x}+ c_2xe^{2x}+ c_3\cos x+c_4\sin x}$$

解答: $$A=\begin{bmatrix}1 & 2 & 1 \\6 & -1 & 0 \\ -1 & -2 & -1 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -\lambda(\lambda+4)(\lambda-3)=0 \Rightarrow \lambda=0,-4,3\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}1 & 2 & 1 \\6 & -1 & 0 \\ -1 & -2 & -1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{13x_1+x_3=0\\ 6x_1=x_2} \\\qquad \Rightarrow v=x_3 \begin{pmatrix}\frac{-1}{13} \\\frac{-6}{13} \\1 \end{pmatrix}, \text{choose }v_1 =\begin{pmatrix} 1 \\ 6 \\-13 \end{pmatrix}\\ \lambda_2=-4 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}5 & 2 & 1 \\6 & 3 & 0 \\ -1 & -2 & 3 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2=2x_3}  \\\qquad \Rightarrow v=x_3 \begin{pmatrix}-1 \\ 2 \\1 \end{pmatrix}, \text{choose }v_2 = \begin{pmatrix}-1 \\ 2 \\1 \end{pmatrix} \\ \lambda_3=3 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}-2 & 2 & 1 \\6 & -4 & 0 \\ -1 & -2 & -4 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1 +x_3=0\\ 2x_2+ 3x_3=0}  \\\qquad \Rightarrow v=x_3 \begin{pmatrix}-1 \\ -3/2 \\1 \end{pmatrix}, \text{choose }v_3 = \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{ \text{ eigenvalues: }0,-4,3, \text{ and eigenvectors: }\begin{pmatrix} 1 \\ 6 \\-13 \end{pmatrix},\begin{pmatrix}-1 \\ 2 \\1 \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \\ -2 \end{pmatrix}}$$

解答: $$L\{y''\}-10L\{ y'\}+9L\{ y\} =5L\{t\} \Rightarrow s^2Y(s)+s-2-10(sY(s)+1)+9Y(s)={5\over s^2} \\ \Rightarrow (s^2-10s+9)Y(s)={5\over s^2}-s+12 \Rightarrow Y(s)={5\over s^2(s-1)(s-9)}-{s-12\over (s-1)(s-9)} \\ ={-2\over s-1}+{31\over 81}\cdot {1\over s-9}+{50\over 81}\cdot {1\over s}+{5\over 9}\cdot {1\over s^2} \\ \Rightarrow y(t)= L^{-1}\{Y(s)\} \Rightarrow \bbox[red, 2pt]{y(t)=-2e^{t}+{31\over 81}e^{9t}+{50\over 81}+{5\over 9}t}$$

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解題僅供參考,其他歷年試題及詳解