國立清華大學 114學年度碩士班考試入學試題
系所班組別:工程與系統科學系碩士班 乙組(0532)
考試科目(代碼):工程數學 (3201)
解答:$$\textbf{(a) }{dy\over dx}+{y\over x}=e^x \Rightarrow x{dy\over dx}+{y }= xe^x \Rightarrow \left( xy \right)'=xe^x \Rightarrow xy= \int xe^x\,dx =xe^x-e^x+c_1 \\\qquad \Rightarrow \bbox[red, 2pt]{y(x)=e^x-{e^x\over x}+{c_1 \over x}} \\\textbf{(b) }{dy\over dx}={-x\over x^2y+2y} ={-x\over (x^2+2)y} \Rightarrow \int y\,dy= \int{-x\over x^2+2}dx \Rightarrow {1\over 2}y^2= -{1\over 2}\ln(x^2+2)+ c_1 \\\qquad \Rightarrow y^2=-\ln(x^2+2)+c_2 \Rightarrow \bbox[red, 2pt]{y(x)= \pm \sqrt{c_2-\ln(x^2+2)}} \\\textbf{(c) } y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow m(m-1)x^m-mx^m-3x^m=0 \\ \qquad \Rightarrow (m^2-2m-3)x^m=0 \Rightarrow (m-3)(m+1)=0 \Rightarrow m=3,-1 \Rightarrow y_h =c_1x^3 +c_2x^{-1} \\\qquad y_p=Ax \Rightarrow y_p'=A \Rightarrow y_p''=0 \Rightarrow -Ax-3Ax= 8x \Rightarrow A=-2 \Rightarrow y_p= -2x\\ \qquad \Rightarrow y=y_h+y_p =c_1x^3+c_2x^{-1}-2x \Rightarrow y'=3c_1x^2-c_2x^{-2}-2 \Rightarrow \cases{y(1) =c_1 +c_2-2 =0\\ y'(1)=3c_1-c_2-2=0} \\\qquad \Rightarrow \cases{c_1=1 \\ c_2=1} \Rightarrow \bbox[red, 2pt]{y(x)= x^3+ x^{-1}-2x}$$
解答:$$L\{{dy\over dt}\} +3L\{y\} +2 L\{\int_0^t y(\tau)\,d\tau\} = L\{u(t-1)\} -L\{u(t-2)\} \\ \Rightarrow sY(s) +3Y(s)+{2Y(s)\over s}={e^{-s} \over s}-{e^{-2s} \over s} \Rightarrow Y(s)={e^{-s}-e^{-2s} \over s^2+3s+2} \\ \Rightarrow y(t)=L^{-s}\{Y(s)\}= L^{-1} \{\left( e^{-s}-e^{-2s} \right) \left( {1\over s+1}-{1\over s+2} \right)\} \\ \Rightarrow \bbox[red, 2pt]{y(t)= \left( e^{-(t-1)}-e^{-2(t-1)} \right)u(t-1)-\left( e^{-(t-2)}-e^{-2(t-2)} \right)u(t-2)}$$
解答:$$\textbf{(a) }\det(M) =0-16+8-16-0+8= \bbox[red, 2pt]{-16} \Rightarrow M^{-1} ={1\over \det(M)} \begin{bmatrix}C_{11} &C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32}\\C_{13} & C_{23} & C_{33}\end{bmatrix} \\\qquad ={1\over -16} \begin{bmatrix}0& -16& 8\\ -2& -8& 2\\ 4& -16& 4 \end{bmatrix} = \bbox[red, 2pt] {\begin{bmatrix}0& 1& -{1\over 2} \\ {1\over 8}&{1\over 2}& -{1\over 8} \\ -{1\over 4}& 1& -{1\over 4} \end{bmatrix}} \\\textbf{(b) }\det(M-\lambda I) = -(\lambda+2) (\lambda-2)(\lambda-4)=0 \Rightarrow \lambda=\pm 2, 4\\ \lambda_1= -1 \Rightarrow (M-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 2 & 4 & -2 \\-1 & 4 & 1 \\ -4 & 4 & 4\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=0} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 1\\0\\1\end{pmatrix}, \text{choose }v_1= \begin{pmatrix} 1\\0\\1\end{pmatrix} \\ \lambda_2=2 \Rightarrow (M-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -2 & 4 & -2 \\-1 & 0 & 1 \\-4 & 4 & 0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 1\\1 \\1\end{pmatrix}, \text{choose }v_2= \begin{pmatrix} 1\\1 \\1\end{pmatrix} \\ \lambda_3=4 \Rightarrow (M-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} -4 & 4 & -2 \\-1 & -2 & 1 \\-4 & 4 & -2\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=0\\ 2x_2=x_3} \\\qquad \Rightarrow v= x_2 \begin{pmatrix} 0\\ 1\\ 2\end{pmatrix}, \text{choose }v_3= \begin{pmatrix} 0\\ 1\\ 2\end{pmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{\pm 2, 4}, \text{ eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix} 1\\0\\1\end{pmatrix} ,\begin{pmatrix} 1\\1 \\1\end{pmatrix},\begin{pmatrix} 0\\ 1\\ 2\end{pmatrix}}$$
解答:$$\textbf{(a) }\text{Using Divergence Theorem: } \unicode{x222F} _s F\cdot n\,dA= \iiint_V (\nabla\cdot F)\,dV = \iiint_V (3x^2+3y^2 +3z^2)\,dV \\\qquad =\iiint_V 3\rho^2\,dV =\int_0^{2\pi} \int_0^{\pi/2} \int_0^{\sqrt{30}} 3\rho^2 (\rho^2\sin \phi)d\rho d\phi d\theta =\int_0^{2\pi} d\theta \cdot \int_0^{\pi/2} \sin \phi\,d\phi \cdot \int_0^{\sqrt{30}} 3\rho^4d\rho \\ \qquad = (2\pi) \cdot 1\cdot (540\sqrt{30}) = \bbox[red, 2pt]{1080\sqrt{30}\pi} \\\textbf{(b) } F=(z^3,x^3,y^2) \Rightarrow \nabla \times F = \begin{vmatrix}i& j& k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\z^3& x^3 & y^3\end{vmatrix} =(3y^2,3z^2,3x^2) \\ \qquad \Rightarrow I=\oint_C F\cdot r' ds = \iint_S (\nabla\times F)\cdot n\,dA = \iint_S (3y^2,3z^2,3x^2)\cdot (1,0,0)\,dA = \iint_S 3y^2\,dA \\ \qquad \cases{x=2\\ y=r\cos \theta\\ z=r\sin \theta} \Rightarrow \cases{3y^2=3r^2\cos^2\theta\\ dA=r\, dr\,d\theta} \Rightarrow I=\int_0^{2\pi} \int_0^4 3r^2\cos^2\theta\cdot r\,dr\,d\theta \\\qquad =3 \left( \int_0^{2\pi} \cos^2\theta\,d\theta \right) \left( \int_0^4 r^3\,dr \right) =3\cdot (\pi)\cdot (64)= \bbox[red, 2pt]{192\pi} \\\textbf{(c) } f(z)={e^z-e^{2z}\over z^2(z+1)} \Rightarrow \cases{z=0: \text{ a pole of order 2} \\z=-1: \text{ a simple pole}} \Rightarrow z=0,-1皆在C內\\ \qquad \text{Res}(f,-1) = \left. {e^z-e^{2z } \over z^2} \right|_{z=-1} =e^{-1}-e^{-2} \\ \qquad \text{Res}(f,0) = \left. {d\over dz} \left( {e^z-e^{2z }\over z+1} \right) \right|_{z=0} = \left. {(e^z-2e^{2z})(z+1)-(e^z-e^{2z}) \over (z+1)^2} \right|_{z=0} =-1 \\ \qquad \Rightarrow \oint_C f(z)\,dz = \bbox[red, 2pt]{2\pi i(e^{-1}-e^{-2}-1)} \\\textbf{(d) } \omega =z+1 \Rightarrow z=\omega-1 \Rightarrow f(z)=(z+2)\cos {1\over z+1} =(\omega+1)\cos{1\over \omega} \\\qquad =(\omega+1) \left( 1-{1\over 2\omega^2}+{1\over 24\omega^4}- \cdots \right) = \left( \omega-{1\over 2\omega}+{1\over 24\omega^3}-\cdots \right) + \left( 1-{1\over 2\omega^2}+{1\over 24\omega^4}- \cdots \right) \\\qquad =\omega+1-{1\over 2\omega}-{1\over 2\omega^2} +{1\over 24\omega^3} +{1\over 24\omega^4}-\cdots \\\qquad =(z+1)+1-{1\over 2(z+1)}-{1\over 2(z+1)^2} +{1\over 24(z+1)^3} +{1\over 24(z+1)^4}-\cdots\\ \qquad \Rightarrow \bbox[red, 2pt] {f(z) =\sum_{n=0}^\infty{(-1)^n \over (2n)!}(z+1)^{1-2n} + \sum_{n=0}^\infty {(-1)^n\over (2n)!}(z+1)^{-2n}}$$
解答:$$u(r, \theta) =R(r) \Theta(\theta) \Rightarrow R''(r) \Theta(\theta)+{1\over r}R'(r) \Theta(\theta)+{1\over r^2}R(r)\Theta''(\theta)=0 \\ \Rightarrow {r^2R''+rR'\over R} =-{\Theta''\over \Theta} =\lambda \\ \text{B.C.: } \cases{u(r,0) =0\\ u(r, \pi/2)=0} \Rightarrow \cases{\Theta(0) =0\\ \Theta(\pi/2)= 0} \\ \lambda =\mu^2 \gt 0 \Rightarrow \Theta''+\lambda\Theta=0 \Rightarrow \Theta(\theta)=A\cos(\mu \theta) +B\sin (\mu\theta) \Rightarrow \cases{ \Theta(0)=A =0\\ \Theta(\pi/2)=B\sin(\mu\pi /2)=0} \\ \Rightarrow {\mu \pi\over 2} =n\pi \Rightarrow \mu =2n \Rightarrow \lambda_n=(2n)^2=4n^2 \Rightarrow \Theta_n(\theta) =\sin(2n\theta),n\in \mathbb N \\ \lambda_n=4n^2 \Rightarrow r^2R''+rR'-\lambda R= r^2R''+rR'-4n^2R=0 \\ 取R=r^k代入上式\Rightarrow (k(k-1)+k-4n^2)r^k=0 \Rightarrow k(k-1)+k-4n^2=0 \Rightarrow k^2-4n^2=0 \\ \Rightarrow k=\pm 2n \Rightarrow R_n(r)= C_nr^{2n}+D_nr^{-2n} \Rightarrow R_n(r)=r^{2n} \\ \Rightarrow u(r,\theta)= \sum_{n=1}^\infty A_nr^{2n} \sin(2n\theta) \Rightarrow u(c,\theta) =f(\theta) = \sum_{n=1}^\infty A_n c^{2n} \sin(2n\theta) \\ \Rightarrow A_nc^{2n}={2\over \pi/2} \int_0^{\pi/2} f(\theta) \sin {n\pi \theta\over \pi/2} \,d\theta ={4\over \pi} \int_0^{\pi/2} f(\tau) \sin(2n \tau)\,d\tau \\ \Rightarrow \bbox[red, 2pt]{u(r,\theta)= \sum_{n= 1}^\infty A_nr^{2n} \sin(2n\theta), A_n= {4\over \pi c^{2n}} \int_0^{\pi/2} f(\tau)\sin(2n\tau)\,d\tau}$$
解答:$$L\{u(x,t)\} =U(x,s) \Rightarrow L\{u_{xx}\} =L\{u_t\} \Rightarrow {\partial^2 U\over \partial x^2}=sU(x,s)-u(x,0) \Rightarrow {d^2U\over dx^2}-sU=-50 \\ \Rightarrow U_h = c_1 e^{\sqrt s x} +c_2 e^{-\sqrt sx} \Rightarrow U_p=A \Rightarrow 0-sA=-50 \Rightarrow A={50\over s} \Rightarrow U_p={50\over s} \\ \Rightarrow U(x,s)= U_h+U_p \Rightarrow U(x,s) =c_1 e^{\sqrt s x} +c_2 e^{-\sqrt sx} +{50\over s}\\ \textbf{B.C.1}:\lim_{x\to \infty} u(x,t)=50 \Rightarrow \lim_{x\to \infty }U(x,s)={50\over s} \;(c_1 =0) \Rightarrow U(x,s)= c_2e^{-\sqrt sx}+{50\over s} \\ \textbf{B.C. 2}:u(0,t)= \begin{cases} 30,& 0\lt t\lt 1\\ 0, &t\ge 1\end{cases} \Rightarrow u(0,t)=30(u(t)-u(t-1)) \\ \qquad \Rightarrow L\{u(0,t)\} = 30 L\{u(t)-u(t-1)\} \Rightarrow U(0,s)={30\over s}-{30e^{-s}\over s} = c_2+{50\over s} \Rightarrow c_2=-{20\over s}-{30e^{-s} \over s} \\ \Rightarrow U(x,s) = \left( -{20\over s}-{30e^{-s} \over s}\right)e^{-\sqrt sx}+{50\over s} ={50\over s}-{20\over s}e^{-x\sqrt s}-{30\over s}e^{-s}e^{-x\sqrt s} \\ \Rightarrow u(x,t) = L^{-1} \left\{{50\over s} \right\} -L^{-1} \left\{ {20\over s}e^{-x\sqrt s}\right\}-L^{-1} \left\{{30\over s}e^{-s}e^{-x\sqrt s} \right\}\\ \Rightarrow \bbox[red, 2pt]{u(x,t)=50-20\text{ erfc} \left( {x\over 2\sqrt t} \right)-30 \text{ erfc} \left( {x\over 2\sqrt{t-1}} \right)u(t-1)}$$
解答:$$\mathcal F\{u(x,t)\} = \hat u(\omega, t) = \int_{-\infty}^\infty u(x,t) e^{-i\omega x} \,dx \Rightarrow \mathcal F\{a^2{\partial^2 u\over \partial x^2} \}= \mathcal F\{{\partial^2 u\over \partial t^2} \} \Rightarrow a^2(-\omega^2 \hat u) ={\partial^2 \hat u\over \partial t^2} \\ \Rightarrow {d^2 \hat u\over dt^2} +a^2 \omega^2 \hat u=0 \Rightarrow \hat u(\omega ,t)= A(\omega) \cos(a \omega t)+ B(\omega)\sin(a \omega t) \\ u(x,0)=f(x) \Rightarrow \mathcal F\{u(x,0)\} =\mathcal F\{ f(x)\} \Rightarrow \hat u(\omega ,0)=\hat f(\omega) \Rightarrow A(\omega) =\hat f(\omega) \cdots(1) \\ u_t(x,0)= 0 \Rightarrow \left. {d\hat u\over dt} \right|_{t =0} = \left. -a\omega A(\omega) \sin(a\omega t)+ a\omega B(\omega) \cos(a\omega t) \right|_{t=0} \Rightarrow B(\omega) =0 \cdots(2) \\\text{By (1) and (2)} \Rightarrow \hat u(\omega , t)=\hat f(\omega) \cos(a\omega t) \Rightarrow u(x,t) = \mathcal F^{-1}\{\hat u(\omega ,t)\} ={1\over 2\pi} \int_{-\infty}^\infty \hat f(\omega) \cos(a\omega t)e^{i\omega x}\, d\omega \\={1\over 2\pi} \int_{-\infty}^\infty \hat f(\omega) \left( {e^{ia\omega t}+ e^{-ia\omega t} \over 2} \right) e^{i\omega x}\, d\omega = {1\over 2} \left( {1\over 2\pi}\int_{-\infty}^\infty \hat f(\omega)e^{i\omega(x+at)} \,d\omega+ \int_{-\infty}^\infty \hat f(\omega)e^{i\omega(x-at)} \,d\omega\right) \\={1\over 2} \left( f(x+at)+f(x-at) \right) \Rightarrow \bbox[red, 2pt]{u(x,t) ={1\over 2} \left( f(x+at)+f(x-at) \right)}$$
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解題僅供參考，碩士班歷年試題及詳解
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