115年台師大機電碩士班-工程數學詳解

 國立臺灣師範大學115學年度碩士班招生考試

科目: 工程數學  適用系所:機電工程學系

【試題1】(14分)

解答:$$\text{integration factor }I(x)= e^{\int (2/x)\,dx} =x^2 \Rightarrow x^2y'+2xy=x^4e^{-x} \Rightarrow \left( x^2y' \right)'=x^4e^{-x} \\ \Rightarrow x^2y= \int x^4e^{-x}\,dx =  -e^{-x}(x^4 + 4x^3 + 12x^2 + 24x + 24) + C  \\ \Rightarrow \bbox[red, 2pt]{y = -e^{-x}\left(x^2 + 4x + 12 + \frac{24}{x} + \frac{24}{x^2}\right) + \frac{C}{x^2}}$$

解答:$$y''-4y'+4y=0 \Rightarrow r^2-4r+4=0 \Rightarrow (r-2)^2=0 \Rightarrow r=2\\ \Rightarrow \bbox[red, 2pt]{\text{ the complementary solution: }y_c=C_1e^{2x}+C_2xe^{2x}} \\y_p=Ax^2 e^{2x} \Rightarrow y_p' = A(2xe^{2x} + 2x^2e^{2x}) \Rightarrow  y_p''  = A(2 + 8x + 4x^2)e^{2x} \\ \Rightarrow y_p''-4y_p'+4y_p=  A(2 + 8x + 4x^2)e^{2x} - 4[A(2x + 2x^2)e^{2x}] + 4[Ax^2e^{2x}] = e^{2x} \\ \Rightarrow  A[(4 - 8 + 4)x^2 + (8 - 8)x + 2] = 1  \Rightarrow 2A=1\Rightarrow A={1\over 2} \Rightarrow \bbox[red, 2pt]{\text{the particular solution: }y_p={1\over 2}x^2e^{2x}} \\ y= y_c+y_p \Rightarrow \bbox[red, 2pt]{\text{the general solution: }y=C_1e^{2x}+C_2xe^{2x} +{1\over 2}x^2e^{2x}}$$

【試題2】(16分)

解答:$$ \mathcal{L}\{f(t)\} = \mathcal{L}\{te^{-3t}\} + 4\mathcal{L}\{u(t-2)\} ={1\over (s-(-3))^2}+4 \cdot {e^{-2s} \over s} = \bbox[red, 2pt]{{1\over (s+3)^2}+ {4e^{-2s}\over s}}$$

解答:$$ \mathcal{L}\{y'\} + \mathcal{L}\{y\} = \mathcal{L}\{u(t-1)\} \Rightarrow sY(s)-y(0)+ Y(s)={e^{-s}\over s} \Rightarrow Y(s) = \frac{e^{-s}}{s(s+1)} \\ \Rightarrow  Y(s) = e^{-s} \left( \frac{1}{s} - \frac{1}{s+1} \right) \Rightarrow y(t)=  \mathcal{L}^{-1}\{Y(s)\} = u(t-1) \left( 1-e^{-(t-1)} \right) \\ \Rightarrow \bbox[red, 2pt]{ y(t) = \begin{cases} 0, & 0 \le t < 1 \\ 1 - e^{-(t-1)}, & t \ge 1 \end{cases} }$$

【試題3】(16分)

解答:$$\textbf{1. }\det(A-\lambda I)= \det \begin{bmatrix} -2-\lambda & 2 & -3 \\ 2 & 1-\lambda & -6 \\ -1 & -2 & -\lambda \end{bmatrix} =  -\lambda^3 - \lambda^2 + 21\lambda + 45 = 0  \\\quad \Rightarrow  (\lambda - 5)(\lambda + 3)^2 = 0  \Rightarrow \bbox[red, 2pt]{\text{ eigenvalues: 5,-3}} \\\quad \lambda_1=5 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow  \begin{bmatrix} -7 & 2 & -3 \\ 2 & -4 & -6 \\ -1 & -2 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \cases{x=-z\\ y=-2z} \\\quad \Rightarrow v= z \begin{bmatrix}-1\\ -2\\1 \end{bmatrix} , \text{ choosing }v_1= \begin{bmatrix}1\\2\\-1 \end{bmatrix} \\ \lambda_2=-3 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix} 1 & 2 & -3 \\ 2 & 4 & -6 \\ -1 & -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \Rightarrow x+2y=3z \\ \quad \Rightarrow v= y \begin{bmatrix}-2\\1\\0 \end{bmatrix} +z \begin{bmatrix}3\\0\\1 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}-2\\1\\0 \end{bmatrix}, v_3= \begin{bmatrix}3\\0\\1 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: }\begin{bmatrix}1\\ 2\\-1 \end{bmatrix},  \begin{bmatrix}-2\\1\\0 \end{bmatrix},   \begin{bmatrix}3\\ 0\\1 \end{bmatrix}} \\ \textbf{2. }f(x)=x^5 - 4x^3 + 8x^2 - x - 1 = (x^2 - x + 18)(x^3 + x^2 - 21x - 45) + (14x^2 + 332x + 809) \\ \quad \Rightarrow f(A)=14A^2+332A+ 809I = 14  \begin{bmatrix} -2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0 \end{bmatrix} \begin{bmatrix} -2 & 2 & -3 \\ 2 & 1 & -6 \\ -1 & -2 & 0 \end{bmatrix} \\+  \begin{bmatrix} -664 & 664 & -996 \\ 664 & 332 & -1992 \\ -332 & -664 & 0 \end{bmatrix}+  \begin{bmatrix} 809 & 0 & 0 \\ 0 & 809 & 0 \\ 0 & 0 & 809 \end{bmatrix} =\bbox[red, 2pt]{ \begin{bmatrix} 299 & 720 & -1080 \\ 720 & 1379 & -2160 \\ -360 & -720 & 1019 \end{bmatrix}}$$

【試題4】(12分)

解答:$$A(w) = \frac{2}{\pi} \int_0^\infty f(v) \cos(wv) \, dv =  \frac{2}{\pi} \int_0^\infty e^{-\alpha v} \cos(wv) \, dv   \\= {2\over \pi}\left[ \frac{e^{-\alpha v}}{(-\alpha)^2 + w^2} (-\alpha \cos(wv) + w \sin(wv)) \right]_0^\infty ={2\over \pi} \cdot {\alpha\over \alpha^2+w^2}  = \frac{2\alpha}{\pi(\alpha^2 + w^2)}  \\ \Rightarrow f(x) = e^{-\alpha x} =\bbox[red, 2pt]{ \frac{2\alpha}{\pi} \int_0^\infty \frac{\cos(wx)}{\alpha^2 + w^2} \, dw }$$

解答:$$ e^{-(3)(3)} = \frac{2(3)}{\pi} \int_0^\infty \frac{\cos(3x)}{x^2 + 3^2} \, dx \Rightarrow e^{-9} = \frac{6}{\pi} \int_0^\infty \frac{\cos 3x}{x^2+9} \, dx \Rightarrow  \int_0^\infty \frac{\cos 3x}{x^2+9} \, dx = \bbox[red, 2pt]{\frac{\pi}{6} e^{-9} }$$

【試題5】(12分)

解答:$$\text{Let }\cases{x_1 = y \\x_2 = y' \\x_3 = y''} \Rightarrow \cases{x_1'=y'=x_2\\ x_2'=y''=x_3\\ x_3'=y'''} \Rightarrow y''' = 6y-11y'+6y'' \Rightarrow x_3'=6x_1-11x_2+6x_3 \\ \Rightarrow \bbox[red, 2pt]{\begin{bmatrix} x_1' \\ x_2' \\ x_3' \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 6 & -11 & 6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} }$$

解答:$$\text{Let }\cases{x_1 = y \\x_2 = y'  } \Rightarrow  x_1'=x_2 \Rightarrow y''=-50y-15y' \Rightarrow x_2'=-50x_1-15x_2 \\ \Rightarrow  \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -50 & -15 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \\ A=\begin{bmatrix} 0 & 1 \\ -50 & -15 \end{bmatrix} \Rightarrow \det(A-\lambda I)= \lambda^2+15\lambda+50= (\lambda+5)(\lambda+10)=0 \\ \Rightarrow \lambda_1=-5, \lambda_2=-10 \\ \lambda_1=-5 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} 5 & 1 \\ -50 & -10 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 5a + b = 0  \\\qquad \text{Choosing $a=1$ yields $b=-5$. Thus, $\mathbf{v}_1 = \begin{bmatrix} 1 \\ -5 \end{bmatrix}$}. \\ \lambda_2=-10 \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow  \begin{bmatrix} 10 & 1 \\ -50 & -5 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 10a + b = 0  \\ \qquad \text{Choosing $a=1$ yields $b=-10$. Thus, $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -10 \end{bmatrix}$.} \\ \Rightarrow  \mathbf{x}(t) = C_1 e^{-5t} \begin{bmatrix} 1 \\ -5 \end{bmatrix} + C_2 e^{-10t} \begin{bmatrix} 1 \\ -10 \end{bmatrix} \\ \text{ Since }y=x_1 \Rightarrow \bbox[red, 2pt]{y(t) = C_1 e^{-5t} + C_2 e^{-10t} }$$

【試題6】(12分)

解答:$$\text{Let }P(x,y)={-y\over x^2+y^2} \text{ and } Q(x,y) ={x\over x^2+ y^2}, \text{ then }\vec F=P(x,y)\vec i+ Q(x,y)\vec j \\ \Rightarrow \nabla \times \vec F= \left( Q_x-P_y \right)\vec k = \left(    \frac{y^2 - x^2}{(x^2+y^2)^2} -\frac{y^2 - x^2}{(x^2+y^2)^2} \right)\vec k =0\vec k=\bbox[red, 2pt]0$$

解答:$$\vec r=x\vec i+y\vec j \Rightarrow \vec n={x\vec i+y\vec j\over \sqrt{x^2+y^2}} =x\vec i+y\vec j\\ \Rightarrow \vec F\cdot \vec n=  \left( \frac{-y}{x^2+y^2}\vec{i} + \frac{x}{x^2+y^2}\vec{j} \right) \cdot (x\vec{i} + y\vec{j})   = \frac{-xy}{x^2+y^2} + \frac{xy}{x^2+y^2}=0 \\ \Rightarrow \oint_C \vec F\cdot \vec n\,ds=\oint_C 0\,ds =\bbox[red, 2pt]0$$

【試題7】(18分)

解答:$$\text{A partial differential equation (PDE) is a mathematical equation that relates an}\\\text{ unknown function of two or more independent variables to its partial derivatives.}\\ \text{Example Application: The Heat Equation. It models the distribution and flow of heat}\\\text{ (or variation in temperature) within a solid medium over time.}$$

解答:$$u(x,t) =X(x) T(t) \Rightarrow XT'= \alpha^2 X''T \Rightarrow {T'\over \alpha^2 T} ={X''\over X}= -\lambda \Rightarrow \cases{X''+\lambda X=0\\ T'+\alpha^2 \lambda T=0} \\ \text{boundary conditions:} \cases{u(0,t)= X(0)T(t)=0\\ u(L,t)=X(L)T(t) =0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \cases{X(0)=c_2=0 \\ X(L)=c_1L+c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow X=0\\ \lambda \lt 0 \Rightarrow \lambda=-k^2(k\gt 0) \Rightarrow X''-k^2X=0\Rightarrow X=c_1e^{kx}+c_2e^{-kx} \Rightarrow X(0)=c_1+c_2=0\\ \qquad \Rightarrow X=c_1e^{kx}-c_1e^{-kx} \Rightarrow X(L)=c_1(e^{kL} -e^{-kL}) =0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \lambda\gt 0 \Rightarrow \lambda=\omega^2 (\omega\ne 0) \Rightarrow  X  = c_1 \cos(\omega x) + c_2 \sin(\omega x) \Rightarrow X(0)=c_1=0 \Rightarrow X=c_2\sin(\omega x) \\\qquad \Rightarrow X(L) =c_2\sin(\omega L)=0 \Rightarrow \omega L=n\pi \Rightarrow \lambda_n= \left( {n\pi \over L} \right)^2 \Rightarrow X_n=\sin{n\pi x\over L}, n=1,2,\dots \\ \Rightarrow T'+\alpha^2 \left( n\pi\over L \right)^2T=0 \Rightarrow  T_n(t) = C_n e^{-\alpha^2 \left(\frac{n\pi}{L}\right)^2 t} \\ \Rightarrow  u(x,t) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha^2 \left(\frac{n\pi}{L}\right)^2 t} \Rightarrow  u(x,0) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right) = f(x)  \\ \Rightarrow  A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx \\ \Rightarrow \bbox[red, 2pt]{u(x,t) = \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha^2 \left(\frac{n\pi}{L}\right)^2 t}, \text{where }A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx }$$

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