113年台科大自動控制碩士班-工程數學詳解

 國立臺灣科技大學113學年度碩士班招生試題

系所組別：自動化及控制研究所碩士班
科目：工程數學

解答： $$y'=xy^2-xy=x(y^2-y) \Rightarrow {1\over y^2-y} dy= x\,dx \Rightarrow \int \left( {1\over y-1}-{1\over y} \right)\,dy = \int x\,dx \\ \Rightarrow \ln(y-1)-\ln y=\ln {y-1\over y}={1\over 2}x^2+c_1 \Rightarrow {y-1\over y}=1-{1\over y}=c_2e^{x^2/2} \\ \Rightarrow {1\over y}=1-c_2e^{x^2/2} \Rightarrow \bbox[red, 2pt]{ y={1\over 1-c_2e^{x^2/2}}}$$

解答： $$y=x^m \Rightarrow \cases{y'=mx^{m-1}\\ y''=m(m-1)x^{m-2}} \Rightarrow x^2y''-5xy'+10y= (m^2-6m+10)x^m=0 \\ \Rightarrow m=3\pm i \Rightarrow y=x^3(c_1 \cos (\ln x)+ c_2\sin (\ln x)) \\\Rightarrow y'= 3x^2(c_1 \cos (\ln x)+ c_2\sin (\ln x))+ x^3(-c_2{1\over x}\sin(\ln x) +c_2{1\over x} \cos(\ln x)) \\ \Rightarrow \cases{y(1)= c_1=4\\ y'(1)=3c_1+c_2=-6 \Rightarrow c_2=-18} \Rightarrow \bbox[red, 2pt]{y=x^3(4\cos(\ln x)-18\sin(\ln x))}$$

解答： $$\cases{3x'-y=2t \\ x'+y'-y=0 } \Rightarrow \cases{3L\{x'\}-L\{y\}=2L\{t\} \\ L\{x'\}+L\{y'\}-L\{y\}} \Rightarrow \cases{3sX(s)-Y(s)=2/s^2 \cdots(1) \\ sX(s)+sY(s)-Y(s)=0 \cdots(2)} \\(1)-3\times(2) \Rightarrow (2-3s)Y(s)={2\over s^2} \Rightarrow Y(s)={2\over s^2(2-3s)} ={3\over 2}\cdot {1\over s}+{1\over s^2}-{3\over 2}\cdot {1\over s-2/3}\\ \Rightarrow y(t)=L^{-1}\{Y(s)\} \Rightarrow \bbox[red, 2pt]{y(t)= {3\over 2}+t-{3\over 2}e^{2t/3}} \\ (2) \Rightarrow X(s)={1-s\over s}Y(s) ={2(1-s)\over s^3(2-3s)} ={3\over 4s}+{1\over 2s^2}+{1\over s^3}-{3\over 4(s-2/3)} \\ \Rightarrow x(t)=L^{-1}\{ X(s)\}  \Rightarrow \bbox[red, 2pt]{x(t)={3\over 4}+{t\over 2}+{t^2\over 2}-{3\over 4}e^{2t/3}}$$

解答： $$y=\sum_{n=0}^\infty a_nx^n \Rightarrow \cases{y'=\sum_{n=0}^\infty na_nx^{n-1} \\ x^2y = \sum_{n=0}^\infty a_nx^{n+2}} \Rightarrow (1-x^2)y= \sum_{n=0}^\infty (a_n-a_{n-2})x^n,a_n=0 ,\forall n\lt 0 \\ \Rightarrow y'+(1-x^2)y=\sum_{n=0}^\infty (a_n-a_{n-2}+(n+1)a_{n+1})x^n =x \\ \Rightarrow \cases{a_0+a_1=0\\ a_1+2a_2=1\\ a_2-a_0+3a_3=0\\ a_n-a_{n-2}+(n+1)a_{n+1}=0,n\ge 2} \Rightarrow\begin{cases} a_1=-a_0\\ a_2={1\over 2}(1+a_0) \\ a_3={1\over 6}(a_0-1)\\ a_4={1\over 24}(1-7a_0)\end{cases} \\ \Rightarrow\bbox[red, 2pt]{ y=a_0-a_0x+{1\over 2}(1+a_0)x^2 -{1\over 6}(1-a_0)x^3+{1\over 24}(1-7a_0)x^4+ \cdots}$$

解答： $$題目的u(a,0)應該是\bbox[cyan,2pt]{u(a,y)} \\u(x,y)=X(x)Y(y) \Rightarrow u_{xx}+u_{yy}=X''Y+XY''=0 \Rightarrow -{Y''\over Y}={X''\over X} = \lambda \\ \cases{u(0,y)=X(0)Y(y)=0 \\ u(a,y)= X(a)Y(y)=0} \Rightarrow \cases{X(0)=0\\ X(a)=0} \\\textbf{Case I }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \cases{X(0)=c_2=0\\ X(a)=c_1a+c_2=0} \\\qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case II }\lambda = \rho^2 \gt 0 \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x}+ c_2e^{-\rho} =0 \\\qquad \Rightarrow \cases{X(0)=c_1+c_2=0\\ X(a)=c_1e^{a\rho } +c_2e^{-a\rho} =0} \Rightarrow c_2=-c_1 \Rightarrow c_1e^{2a\rho}-c_1= c_1(e^{2a\rho}-1)=0 \\ \qquad \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case III }\lambda=-\rho^2 \lt 0 \Rightarrow X''+\rho^2 X=0 \Rightarrow X=c_1 \cos(\rho x)+ c_2 \sin(\rho x) \\\qquad \Rightarrow X(0)=c_1=0 \Rightarrow X(a)= c_2\sin(a\rho )=0 \Rightarrow \sin(a\rho )=0 \Rightarrow a\rho = n\pi \\\qquad \Rightarrow \rho ={n\pi\over a} \Rightarrow X= \sin({n\pi x\over a}),n=1,2,\dots \\ -{Y''\over Y}=\lambda =-\rho^2 \Rightarrow Y''-\rho^2 Y=0 \Rightarrow Y=c_1\cosh{n\pi y\over a}+ c_2\sinh{n\pi y\over a}\\ u(x,0)= X(x)Y(0)=0 \Rightarrow Y(0)=0 \Rightarrow c_1=0 \Rightarrow Y=c_2\sinh{n\pi y\over a} \\ \Rightarrow \bbox[red, 2pt]{u(x,y)= \sum_{n=1}^\infty A_n \sinh{n\pi y\over a}\sin({n\pi x\over a})} \Rightarrow u(x,b)= f(x)=\sum_{n=1}^\infty A_n \sinh{n\pi b\over a} \sin({n\pi x\over a}) \\ \Rightarrow A_n  \sinh{n\pi b\over a}= {2 \over a}\int_0^a  f(x)\sin({n\pi x\over a})\,dx \Rightarrow \bbox[red, 2pt]{A_n={2 \over a  \sinh{n\pi b\over a}}\int_0^a  f(x)\sin({n\pi x\over a})\,dx}$$

解答： $$A=\begin{bmatrix}0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-(\lambda+1)^2(\lambda-2)=0 \Rightarrow \text{ eigenvalue }\lambda=-1,2 \\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix}=0 \Rightarrow x_1+x_2+x_3=0\\ \qquad \Rightarrow v=x_2 \begin{pmatrix}-1   \\1\\0 \end{pmatrix} +x_3 \begin{pmatrix}-1   \\0\\1 \end{pmatrix}, \text{ choose }v_1=\begin{pmatrix}-1   \\1\\0 \end{pmatrix}, v_2= \begin{pmatrix}-1   \\0\\1 \end{pmatrix} \\\lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2 & 1 & 1 \\1 & -2 & 1 \\1 & 1 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 1  \\1\\1 \end{pmatrix}, \text{ choose }v_3=  \begin{pmatrix} 1  \\1\\1 \end{pmatrix}\\ \Rightarrow P=[v_1 \mid v_2 \mid v_3] =\begin{bmatrix}-1 & -1& 1 \\1 & 0& 1\\ 0 & 1& 1 \end{bmatrix} \Rightarrow P^{-1}= \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{ A=\begin{bmatrix}-1 & -1& 1 \\1 & 0& 1\\ 0 & 1& 1 \end{bmatrix} \begin{bmatrix}-1 & 0& 0 \\0 & -1 & 0\\ 0 & 0 & 1 \end{bmatrix}  \begin{bmatrix} \frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{bmatrix}}$$

解答： $$a_0={1\over 2\pi} \int_0^h 1\,dt ={h\over 2\pi}\\ a_n= {1\over \pi} \int_0^h  \cos(nt)\,dt ={1\over \pi}\left( \left. \left[{1\over n}\sin(nt) \right] \right|_0^h\right) ={1\over n\pi} \sin(nh),n\in \mathbb N \\ b_n= {1\over \pi}\int_0^h \sin(nt)\,dt ={1\over n\pi}(1-\cos(nh)), n\in \mathbb N \\ f(t)=a_0+ \sum_{n=1}^\infty (a_n\cos(nt)+b_n\sin(nt) ) \\={h\over 2\pi} + \sum_{n=1}^\infty \left( {1\over n\pi} \sin(nh)\cos(nt)+{1\over n\pi}(1-\cos(nh)) \sin(nt) \right) \\\Rightarrow\bbox[red, 2pt]{ f(t)={h\over 2\pi^2}+ \sum_{n=1}^\infty{1\over n}\left( \sin(nh-nt) + \sin(nt)\right) }$$

解答： $$F(\omega)=\int_{-\infty}^\infty f(x)e^{-j\omega x}\,dx =\int_{-a}^a 2e^{-j\omega x}\,dx =\left. \left[ {2\over -j\omega} e^{-j\omega x} \right] \right|_{-a}^a ={2\over -j\omega}\left( e^{-j\omega a} -e^{j\omega a}\right) \\\qquad ={2\over j\omega} (e^{j\omega a}-e^{-j \omega a})={2\over j\omega}\cdot 2\sin(\omega a)\cdot j =\bbox[red, 2pt]{4\sin(\omega a) \over \omega}$$

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