110年北科大土木碩士班-工程數學詳解

 國立臺北科技大學110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
科目:工程數學

解答: $$\textbf {1.}\; y'={3x^2\over 2y} \Rightarrow 2ydy = 3x^2dx \Rightarrow y^2=x^3+c_1\\\quad y(1)=3 \Rightarrow 3^2=1+c_1 \Rightarrow c_1=8 \Rightarrow \bbox[red, 2pt]{y= \sqrt{x^3+8}} \\\textbf{2.}\; y'=xe^x \Rightarrow y=\int xe^x\,dx =xe^x-e^x+c_1 \Rightarrow \bbox[red, 2pt]{y=xe^x-e^x+c_1} \\\textbf{3.}\; y''+4y'-y=0 \Rightarrow \lambda^2+4\lambda-1=0 \Rightarrow \lambda=-2\pm \sqrt 5 \Rightarrow y_h=c_1e^{(-2+\sqrt 5)x} +c_2e^{-(2+\sqrt 5)x} \\ \quad y_p=ax^2+bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p''=2a \\\quad \Rightarrow y_p''+4y_p'-y_p= -ax^2+(8a-b)x+(2a+4b-c) =2x^2-3x+6 \\ \quad \Rightarrow \cases{-a=2\\ 8a-b=-3\\ 2a+4b-c=6} \Rightarrow \cases{a=-2\\ b=-13\\ c=-62} \Rightarrow y_= -2x^2-13x-62\\\quad \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{(-2+\sqrt 5)x} +c_2e^{-(2+\sqrt 5)x}-2x^2-13x-62} \\ \textbf{4.} \; y'''+6y''+11y'+6y=0 \Rightarrow \lambda^3+6\lambda^2+11\lambda +6=0 \Rightarrow (\lambda+1)(\lambda +2)(\lambda +3)=0 \\\quad \Rightarrow \lambda=-1,-2,-3 \Rightarrow  \bbox[red, 2pt]{y= c_1e^{-x}+ c_2e^{-2x} +c_3e^{-3x}}$$

解答: $$A=\begin{bmatrix}5 & 10 & -10 \\10 & 5 & 10 \\-10 & 10 & 5 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-(\lambda+15)(\lambda-15)^2=0 \Rightarrow \lambda=-15,15\\ \lambda_1=-15 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix}20 & 10 & -10 \\10 & 20 & 10 \\-10 & 10 & 20\end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2+x_3=0} \\ \qquad \Rightarrow \mathbf x= k\left(\begin{array}{c}1\\ -1 \\1 \end{array} \right), k\in \mathbb R, \text{choose }v_1= \left(\begin{array}{c}1\\ -1 \\1 \end{array} \right) \\ \lambda_2=15 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix}-10 & 10 & -10 \\10 & -10 & 10 \\-10 & 10 & -10 \end{bmatrix}  \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1-x_2+x_3=0 \\\qquad \Rightarrow \mathbf x =s\begin{pmatrix}1 \\1\\ 0 \end{pmatrix} +t\begin{pmatrix}-1 \\0\\ 1 \end{pmatrix}, \text{choose }v_2=\begin{pmatrix}1 \\1\\ 0 \end{pmatrix}, v_3= \begin{pmatrix} -1 \\0 \\ 1 \end{pmatrix}\\ \Rightarrow \text{eigenvectors: }  \bbox[red,2pt]{-15,15}, \text{ eigenvectors: }\bbox[red,2pt]{\left( \begin{array}{c}1\\ -1 \\1\end{array} \right), \begin{pmatrix}1 \\1\\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\0 \\ 1 \end{pmatrix}}$$

解答: $$L\{y''\}+ L\{y'\}=L\{\pi\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)={\pi \over s} \\ \Rightarrow (s^2+s)Y(s)={\pi\over s}+\pi(s+1) \Rightarrow Y(s)= {\pi \over s^2(s+1)} +{\pi\over s} \\ \Rightarrow y(t)=L^{-1}\{Y(s)\} = L^{-1}\left\{  {\pi \over s^2(s+1)} \right\} + L^{-1}\left\{ {\pi\over s}\right\}\\ = L^{-1}\left\{  -{\pi \over s }+{\pi\over s^2}+{\pi\over s+1} \right\} + L^{-1}\left\{ {\pi\over s}\right\} =-\pi+\pi t+\pi e^{-t}+\pi \Rightarrow \bbox[red, 2pt]{y(t)=\pi(t+e^{-t}) }$$

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