113年台北科大車輛工程碩士班-工程數學詳解

 國立臺北科技大學113學年度碩士班招生考試

系所組別: 車輛工程系碩士班
第一節 工程數學

解答: $$\textbf{1.}\; y'=-{4x\over y} \Rightarrow \int y\,dy =-\int 4x\,dx \Rightarrow {1\over 2}y^2=-2x^2+c_1 \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-4x^2+c_2}} \\ \textbf{2.}\; y(2)=3 \Rightarrow 3=\sqrt{-16+c_2} \Rightarrow c_2=25 \Rightarrow  \bbox[red, 2pt]{y= \sqrt{25-4x^2}}$$

解答: $$\textbf{1.}\; y'=-{4x\over y} \Rightarrow \int y\,dy =-\int 4x\,dx \Rightarrow {1\over 2}y^2=-2x^2+c_1 \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-4x^2+c_2}} \\ \textbf{2.}\; y(2)=3 \Rightarrow 3=\sqrt{-16+c_2} \Rightarrow c_2=25 \Rightarrow  \bbox[red, 2pt]{y= \sqrt{25-4x^2}}$$

解答: $$\textbf{1.}\; y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda = \pm i \Rightarrow \bbox[red, 2pt]{y_h =c_1\cos x+c_2\sin x} \\ \textbf{2.}\; \cases{y_1=\cos x\\ y_2=\sin x} \Rightarrow W=\begin{vmatrix}\cos x & \sin x \\-\sin x & \cos x \end{vmatrix} =1. \text{ Applying variations of parameters,} \\ y_p=-\cos x\int \sin x(\cos x-\sin x)\,dx +\sin x \int \cos x(\cos x-\sin x)\,dx \\\qquad =-\cos x \left(-{1\over 4}\cos(2x)-{1\over 2}x+{1\over 4} \sin (2x) \right)+ \sin x \left( {1\over 4}\sin (2x)+{1\over 2}x +{1\over 4} \cos (2x)\right) \\\qquad ={1\over 4}\left( \cos x-\sin x\right)+{1\over 2}x( \cos x+\sin x) \Rightarrow \bbox[red, 2pt]{y_p={1\over 4}\left( \cos x-\sin x\right)+{1\over 2}x( \cos x+\sin x)}\\ \textbf{3.}\; y=y_h+y_p \Rightarrow \bbox[red, 2pt] {y=c_3\cos x+c_4\sin x +{1\over 2}x( \cos x+\sin x)}$$

解答: $$\textbf{1.}\; L\{f(t)\} =L\{ \cos(\omega t+\theta) \}=  L\{ \cos(\omega t) \cos \theta - \sin(\omega t) \sin \theta\} \\\qquad = \cos \theta L\{\cos \omega t\} -\sin \theta L\{ \sin(\omega t)\} =\cos \theta\cdot { s \over s^2+\omega^2} -\sin \theta \cdot {\omega \over s^2+ \omega^2} =\bbox[red, 2pt]{s\cos \theta-\omega \sin \theta \over s^2+\omega^2} \\\textbf{2.}\; F(s)={s+1\over s^2+9} = {s\over s^2+3^2}+{1\over 3}\cdot {3\over s^2+3^2} \\\quad \Rightarrow L^{-1}\{ F(s)\} =L^{-1} \left\{{s\over s^2+3^2} \right\} +{1\over 3}L^{-1} \left\{ {3\over s^2+3^2} \right\} = \bbox[red, 2pt]{\cos(3t)+{1\over 3}\sin(3t)}$$

解答: $$\textbf{1.}\; [A\mid I]= \left[ \begin{array}{rrr| rrr} -1 & 1 & 2 & 1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-1 & 3 & 4 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_2+3R_1 \to R_2, R_3+R_1\to R_3} \left[ \begin{array}{rrr| rrr} -1 & 1 & 2 & 1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 2 & 2 & -1 & 0 & 1 \end{array} \right]  \\ \xrightarrow{R_2/2\to R_2, R_3/2\to R_3}\left[ \begin{array}{rrr| rrr} -1 & 1 & 2 & 1 & 0 & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 1 & 1 & - \frac{1}{2} & 0 & \frac{1}{2} \end{array} \right] \xrightarrow{R_1-R_2\to R_1,R_3-R_2\to R_3} \left[ \begin{array}{rrr| rrr}  -1 & 0 & - \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & - \frac{5}{2} & -2 & - \frac{1}{2} & \frac{1}{2}\end{array} \right]  \\ \xrightarrow{-(2/5)R_3\to R_3} \left[ \begin{array}{rrr| rrr} -1 & 0 & - \frac{3}{2} & - \frac{1}{2} & - \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right] \xrightarrow{R_1+ (3/2)R_3 \to R_2, R_2-(7/2)R_3 \to R_2}\\ \left[ \begin{array}{rrr| rrr} -1 & 0 & 0 & \frac{7}{10} & - \frac{1}{5} & - \frac{3}{10}\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right] \xrightarrow{-R_1 \to R_1} \left[ \begin{array}{rrr| rrr}  1 & 0 & 0 & - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \\ \Rightarrow A^{-1}= \bbox[red,2pt]{ \begin{bmatrix}- \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\- \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\\frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{bmatrix}} \\ \textbf{2.}\; A=\begin{bmatrix}1 & -1 & 0 \\-1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-\lambda(\lambda-1)(\lambda-3) =0 \Rightarrow \lambda=0,1,3\\ \lambda_1= 0 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & -1 & 0 \\-1 & 2 & -1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \\ \qquad \Rightarrow v=x_3 \begin{pmatrix}1 \\1\\ 1 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix}1 \\1\\ 1 \end{pmatrix} \\\lambda_2=1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}0 & -1 & 0 \\-1 & 1 & -1 \\0 & -1 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_2=0\\ x_1+x_3=0} \\\qquad \Rightarrow v= x_3 \begin{pmatrix}-1 \\0 \\ 1 \end{pmatrix}, \text{choose }v_2= \begin{pmatrix}-1 \\0 \\ 1 \end{pmatrix} \\ \lambda_3=3  \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} -2 & -1 & 0 \\-1 & -1 & -1 \\0 & -1 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2+2x_3=0} \\ \qquad \Rightarrow v=x_3 \begin{pmatrix} 1 \\-2 \\ 1 \end{pmatrix}, \text{choose }v_3= \begin{pmatrix} 1 \\-2 \\ 1 \end{pmatrix}\\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{0,1,3}, \text{ and eigenvectors: } \bbox[red, 2pt]{ \begin{pmatrix}1 \\1\\ 1 \end{pmatrix},  \begin{pmatrix}-1 \\0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\-2 \\ 1 \end{pmatrix}}$$

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

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