114年台科大自動碩士班-工程數學詳解

國立臺灣科技大學114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:$$y'+3y=0 \Rightarrow y_h= c_1e^{-3x}\\ y_p=Ae^{2x}+B \Rightarrow y_p'=2Ae^{2x} \Rightarrow y_p'+3y_p=5Ae^{2x}+3B=5e^{2x}-6 \Rightarrow \cases{A=1\\ B=-2} \\ \Rightarrow y_p=e^{2x}-2 \Rightarrow y=y_h+ y_p=c_1e^{-3x}+e^{2x}-2 \Rightarrow y(0)=c_1+1-2=2 \Rightarrow c_1=3\\ \Rightarrow \bbox[red, 2pt]{y=3e^{-3x}+e^{2x}-2}$$

解答:$$y''-2y'+10y=0 \Rightarrow \lambda^2-2\lambda+10=0 \Rightarrow \lambda=1\pm 3i  \Rightarrow \bbox[red, 2pt]{y=e^x(c_1\cos(3x)+ c_2\sin(3x))}$$

解答:$$f(t+6)=f(t) \Rightarrow L\{f(t)\} ={ \int_0^6 e^{-st}f(t)\,dt\over 1-e^{-6s}} ={1\over 1-e^{-6s}}\int_0^3 5e^{-st}\,dt ={5\over 1-e^{-6s}} \left. \left[ -{1\over s}e^{-st}\right]\right|_0^3 \\= {5\over 1-e^{-6s}}\left(-{1\over s}e^{-3s}+{1\over s} \right) ={5\over s(1-e^{-6s})} \left( 1-e^{-3s} \right) = \bbox[red, 2pt]{5\over s(1+e^{-3s})}$$

解答:$$y= \sum_{n=0}^\infty a_nx^n \Rightarrow y'=\sum_{n=0}^\infty na_nx^{n-1} \Rightarrow y'-xy=\sum_{n=0}^\infty na_nx^{n-1} -\sum_{n=0}^\infty a_nx^{n+1} \\=a_1+ \sum_{n=2}^\infty (na_n-a_{n-2})x^{n-1} =1-x \Rightarrow \cases{a_1=1\\ 2a_2-a_0=-1\\ na_n-a_{n-2}=0, n\ge 3} \\ \Rightarrow \bbox[red, 2pt]{y=(2a_2+1)+x +a_2x^2 +{1\over 3}x^3+{1\over 4}a_2x^4+{1\over 15}x^5+\cdots}$$

解答:$$5\mathbf i-12\mathbf k={17\over 2}(2\mathbf i)+4(3\mathbf j)-12(\mathbf i+\mathbf j+\mathbf k) \Rightarrow \bbox[red, 2pt]{\text{linearly dependent}}$$

解答:$$\textbf{(1) }A= \begin{bmatrix} 1&-3\\ 2&-4\end{bmatrix} \Rightarrow \det(A-\lambda I) =(\lambda+2)(\lambda+1)=0 \Rightarrow \lambda=-2,-1\\ \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 3 & -3 \\2 & -2\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=x_2 \Rightarrow v=x_2 \begin{pmatrix} 1\\ 1\end{pmatrix} \\\qquad \Rightarrow \text{choose }v_1=\begin{pmatrix} 1\\ 1\end{pmatrix} \\\lambda_2=-1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix} 2 & -3 \\2 &-3 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow 2x_1=3x_2 \Rightarrow v=x_2 \begin{pmatrix} 3/2\\ 1\end{pmatrix} \\\qquad \Rightarrow \text{choose }v_2 =\begin{pmatrix} 3/2\\ 1\end{pmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{-2,-1} \text{ and the corresponding eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix} 1\\ 1\end{pmatrix}, \begin{pmatrix} 3/2\\ 1\end{pmatrix}} \\\textbf{(2) }p(\lambda) =\det(A-\lambda I)=\lambda^2+3\lambda+2 \Rightarrow p(A)=0 \Rightarrow A^2+3A+2I=0 \\ x^{50} =Q(x)p(x)+ax+b \Rightarrow \cases{(-2)^{50}= -2a+b \\ (-1)^{50} =-a+b} \Rightarrow \cases{a=1-2^{50}\\b=2-2^{50}} \\ \Rightarrow A^{50}=aA+bI=(1-2^{50})A+(2-2^{50})I =\begin{bmatrix}(1-2^{50})& -3(1-2^{50})\\ 2(1-2^{50})& -4(1-2^{50}) \end{bmatrix}+ \begin{bmatrix}2-2^{50}& 0\\ 0 &2-2^{50} \end{bmatrix}\\ = \bbox[red, 2pt]{\begin{bmatrix}3-2^{51}& -3+3\cdot 2^{50}\\ 2-2^{51}& -2+3\cdot 2^{51} \end{bmatrix}}$$

解答:$$a_0= {1\over 2}\int_0^2 (1-x^3)\,dx = -1\\ a_n= \int_0^2 (1-x^3) \cos{n\pi x\over 2}\,dx = -{48\over n^2\pi^2}(-1)^{n }-{96\over n^4\pi^4} \left( 1-(-1)^{n} \right) \\ \Rightarrow \bbox[red, 2pt]{\text{Fourier cosine series: }f(x)\sim -1 -\sum_{n=1}^\infty \left[ {48\over n^2\pi^2}(-1)^{n }+{96\over n^4\pi^4} \left( 1-(-1)^{n} \right) \right] \cos{n\pi x\over 2}} \\ b_n= \int_0^2 (1-x^3) \sin {n\pi x\over 2}\,dx = {2+14(-1)^n\over n\pi}-{96(-1)^n\over n^3\pi^3} \\ \Rightarrow \bbox[red, 2pt]{ \text{Fourier sine series: }f(x) \sim \sum_{n=1}^\infty \left[{2+14(-1)^n\over n\pi}-{96(-1)^n\over n^3\pi^3} \right] \sin {n\pi x\over 2} }\\ \Rightarrow  \bbox[red, 2pt]{\text{Sum of the series: }f(x)=\begin{cases}1-x^3,& 0\lt x\lt 2\\ 0,& x=0,2 \end{cases}}$$

解答:$$\mathcal F\{y''\} +6\mathcal F\{y'\} +5\mathcal F\{y\} =\mathcal F\{\delta(t-3)\} \Rightarrow (-\omega^2) Y(\omega) +6(i\omega)Y(\omega)+5 Y(\omega)=e^{-i3\omega} \\ \Rightarrow Y(\omega) ={e^{-i3\omega} \over -\omega^2+ 6i\omega+5} =-{e^{-i2\omega} \over (\omega-3i)^2+4} \Rightarrow \mathcal F^{-1} \{Y(\omega)\} =-{1\over 4}e^{3(t-3)}e^{-2|t|} \\ \Rightarrow \bbox[red, 2pt]{y(t)=-{1\over 4}e^{3(t-3)}e^{-2|t|}}$$

解答:$$z=x+i y \Rightarrow f(z)=(x+iy)^2-i(x+iy) =(x^2-y^2+y)+i(2xy-x) \Rightarrow \cases{u(x,y)=x^2-y^2+y\\ v(x,y)=2xy-x} \\ \Rightarrow \cases{u_x=2x\\ u_y=-2y+1\\ v_x=2y-1\\ v_y=2x} \Rightarrow \cases{u_x=v_y\\ u_y=-v_x} \Rightarrow \bbox[red, 2pt]{\text{Yes, the Cauchy-Riemann equations hold for all }(x,y)} \\ \bbox[red, 2pt]{f(z)=z^2 -iz \text{ is differentiable for all }z\in \mathbb C}$$

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解題僅供參考，碩士班歷年試題及詳解