113年台北科大機械碩士甲-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: 機械工程系機電整合碩士班甲組
第一節 工程數學

解答: $$\textbf{1.}\; \cases{P(x,y)=2xy\\ Q(x,y)= x^2-y^2} \Rightarrow P_y=2x=Q_x \Rightarrow \text{exact} \\ \quad \Rightarrow \Phi(x,y)= \int 2xy\,dx = \int (x^2-y^2) \,dy \Rightarrow \Phi=x^2y+ \phi(y)=x^2y-{1\over 3}y^3+ \rho(x) \\ \quad \Rightarrow \Phi= \bbox[red, 2pt]{x^2y-{1\over 3}y^3+c_1 =0} \\ \textbf{2.}\; xy'-y=y^2 \Rightarrow xdy=(y^2+y)dx \Rightarrow (y^2+y)\,dx+ (-x)\,dy=0 \\\quad \cases{P(x,y)=y^2+y\\ Q(x,y)=-x} \Rightarrow \cases{P_y=2y+1 \\ Q_x=-1} \Rightarrow P_y\ne Q_x \Rightarrow \text{Not exact} \\\quad {-(P_y-Q_x )\over P} ={-(2y+2) \over y^2+y} =-{2\over y} \text{ depentent on }y \\ \Rightarrow u'=-{2\over y}u \Rightarrow \text{integrating factor }u(y)={1\over y^2} \Rightarrow \cases{uP=1+1/y\\ uQ=-x/y^2} \\\quad \Rightarrow (uP)_y=-{1\over y^2}=(uQ)_x \Rightarrow \text{exact}  \Rightarrow \Phi(x,y)= \int (1+{1\over y})\,dx = \int -{x\over y^2}\,dy \\\quad \Rightarrow \Phi= x+{x\over y}+ \phi(y)= {x\over y} + \rho(x) \Rightarrow \Phi= x+{x\over y}=c_1 \Rightarrow \bbox[red, 2pt]{y={x\over c_1-x}}$$

解答: $$\textbf{1.}\; y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \quad \Rightarrow x^2y''-2xy'-4y= (m^2-3m-4)x^m=0 \Rightarrow m^2-3m-4=(m-4)(m+1)=0\\ \quad \Rightarrow m=4,-1 \Rightarrow \bbox[red, 2pt]{y=c_1x^4+c_2 x^{-1}} \\\textbf{2.}\; y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1= (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow y_h=c_1e^x +c_2xe^x\\\quad y_p=Ax^2e^x \Rightarrow y_p'=2Axe^x+Ax^2e^x \Rightarrow y_p''=2Ae^x+4Axe^x+Ax^2e^x \\ \quad \Rightarrow y_p''-2y_p'+y_p= 2Ae^x = 2e^x \Rightarrow A=1 \Rightarrow y_p = x^2e^x \\ \quad \Rightarrow y= y_h+y_p \Rightarrow \bbox[red, 2pt]{ y=c_1e^x +c_2xe^x+ x^2e^x}$$

解答: $$\textbf{1.}\; \bbox[cyan,2pt]{題目有誤}\\\quad F(s)=t\cos 3t 變數是t不是s\\ \textbf{2.}\; L\{y''\} +3L\{y'\}+2L\{y \}=2L\{u(t-1)\} \Rightarrow s^2Y(s)+3sY(s)+2Y(s)={2e^{-s} \over s} \\\quad \Rightarrow Y(s)= {2e^{-s} \over s(s+1)(s+2)} =e^{-s}\left( {1\over s}-{2\over s+1}+{1 \over s+2}\right) \\\quad \Rightarrow y(t)=L^{-1}\{Y(s)\} =u(t-1) \left(1-2e^{-(t-1)}+e^{-2(t-1)} \right) \\\quad \Rightarrow \bbox[red, 2pt]{y(t)=u(t-1) \left(1-2e^{-t+1}+e^{-2(t-1)} \right) }$$

解答: $$\textbf{1.}\; A=\begin{bmatrix}10 & -18 \\6 & -11 \end{bmatrix} \Rightarrow \det(A-\lambda I)= (\lambda+2)(\lambda-1)=0 \Rightarrow \lambda=-2,1 \\\quad \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}12 & -18 \\6 & -9 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =0 \Rightarrow 2x_1=3x_2 \\\qquad \Rightarrow v= x_2 \begin{pmatrix}3/2 \\1 \end{pmatrix}, \text{choose }v_1=\begin{pmatrix}3/2 \\1 \end{pmatrix} \\ \lambda_2=1  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}9 & -18 \\6 & -12 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =0 \Rightarrow x_1=2x_2 \\\qquad \Rightarrow v= x_2 \begin{pmatrix}2 \\1 \end{pmatrix}, \text{choose }v_2=  \begin{pmatrix}2 \\1 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{特徵值:-2,1, 特徵向量: \begin{pmatrix}3/2 \\1 \end{pmatrix}, \begin{pmatrix}2 \\1 \end{pmatrix}} \\\textbf{2.}\; \text{Let }P=[v_1\mid v_2], D=\begin{bmatrix}\lambda_1 & 0 \\0 & \lambda_2 \end{bmatrix}, \text{ then } A= PDP^{-1} = \bbox[red, 2pt]{\begin{bmatrix}\frac{3}{2} & 2 \\1 & 1 \end{bmatrix} \begin{bmatrix}-2 & 0 \\0 & 1 \end{bmatrix}\begin{bmatrix}-2 & 4 \\2 & -3 \end{bmatrix}} \\ \textbf{3.}\; e^A= \begin{bmatrix}\frac{3}{2} & 2 \\1 & 1 \end{bmatrix} \begin{bmatrix}e^{-2} & 0 \\0 & e \end{bmatrix} \begin{bmatrix}-2 & 4 \\2 & -3 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} {4e-3e^{-2}}  &  {-6e+6e^{-2}}  \\ {2e-2e^{-2}}  & {-3e+4e^{-2}}  \end{bmatrix}}$$

解答: $$\textbf{1.}\; f(-t)=-f(t) \Rightarrow f\text{ is odd} \Rightarrow a_n=0 \\\quad b_n= {1\over 2} \left(\int_{-2}^0 -2k \sin{n\pi x\over 2}\,dx + \int_0^2 2k \sin{n\pi x\over 2}\,dx\right)\\\qquad ={k\over 2}\left( \left. \left[ {4 \over n\pi} \cos{n\pi x\over 2} \right] \right|_{-2}^0 + \left. \left[ -{4\over n\pi} \cos{n\pi x\over 2}\right] \right|_{0}^2\right) ={k\over 2}\cdot {8\over n\pi} (1-(-1)^n) ={4\over n\pi}(1-(-1)^n) \\ \Rightarrow \mathcal F(f(t))= \bbox[red, 2pt]{ \sum_{n=1}^\infty {4k\over n\pi}(1-(-1)^n) \sin{n\pi x\over 2}} \\  \textbf{2.}\; g(t)=e^{-|t|} \Rightarrow \hat g(\omega)=\int_{-\infty}^\infty g(t)e^{-i\omega t}\,dt = {2\over 1+\omega^2} \\ \quad h(t)=e^{-|t-1|} =g(t-1) = e^{-i\omega} \hat g(\omega)  ={2\over 1+\omega^2}e^{-i\omega} \\ \quad f(t)=2g(t)+h(t) \Rightarrow \hat f(\omega)= 2\hat g(\omega)+e^{-i\omega}\hat g(\omega) \Rightarrow \hat f(\omega)= \bbox[red, 2pt]{{4\over 1+\omega^2} + {2\over 1+\omega^2}e^{-i\omega}}$$
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