113年中山海工碩士班-工程數學詳解

 國立中山大學113學年度碩士班招生考試

科目名稱: 工程數學【離岸風電碩士班、海工系碩士班甲組、海工聯合碩士班】

解答: $$\textbf{(a)}\; y=x^m \Rightarrow \cases{y'=mx^{m-1}\\ y''=m(m-1)x^{m-2}} \Rightarrow x^2y''-3xy'+4y=(m^2-4m+4)x^m=0 \\ \quad \Rightarrow (m-2)^2=0 \Rightarrow m=2 \Rightarrow y_h= c_1x^2 +c_2x^2\ln x \\ \quad y_p =A\ln x+Bx+C \Rightarrow y_p'={A\over x}+B \Rightarrow y_p''=-{A\over x^2} \\\Rightarrow x^2y_p'' -3xy_p'+4y_p= 4A\ln x+Bx-4A+4C=2x+4\ln x \Rightarrow \cases{A=1\\ B=2\\ C=1} \\ \Rightarrow y_p = \ln x+2x+1 \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1x^2 +c_2x^2\ln x+  \ln x+2x+1} \\\textbf{(b)}\;y=\sum_{n=0}^\infty a_n x^n \Rightarrow y'=\sum_{n=0}^\infty na_n x^{n-1} \Rightarrow y''= \sum_{n= 0}^\infty n(n-1)a_n x^{n-2}\Rightarrow \cases{y(0)=a_0=1\\ y'(0)= a_1=1} \\ \Rightarrow y''+4xy'+2y= \sum_{n=0}^\infty ((n+2)(n+1)a_{n+2}+(4n+1)a_n) x^n =0 \\ \Rightarrow  a_{n+2}=-{4n+1\over (n+2)(n+1)}a_n,a_0=a_1=1, n=0,1,2,\cdots$$

解答: $$\int_C \vec F\cdot d\vec r = \iint_S \text{curl }\vec F\cdot \vec n\,dS\\ \text{L.H.S }\\z=2 \Rightarrow C:x^2+y^2=4 \Rightarrow \cases{x(t)=2\cos t\\ y(t)=2\sin t \\z(t)=2} \Rightarrow \cases{x'(t)=-2\sin tdt\\ y'(t)=2\cos t\,dt \\z'(t)=0},0\le t\le 2\pi\\ \vec F=(3y,-xz,yz^2)=(6\sin t,-4\cos t,8\sin t)\\  \Rightarrow \int_C \vec F\cdot d\vec r= \int_0^{2\pi} (6\sin t,-4\cos t,8\sin t) (-2\sin t,2\cos t,0)\,dt = \int_0^{2\pi}(-12\sin^2 t-8\cos^2 t)\,dt\\ \text{Since ,the surface S lies below the curve }C \Rightarrow \int_C \vec F\cdot d\vec r= \int_0^{2\pi}(12\sin^2 t+ 8\cos^2 t)\,dt \\=\int_0^{2\pi} (8+4\sin^2 t)\,dt =\int_0^{2\pi} (8+2-2\cos 2 t)\,dt =\int_0^{2\pi} 10\,dt =20\pi \\ \text{R.H.S}\\\vec F=(3y,-xz,yz^2) \Rightarrow \text{curl }\vec F=\begin{bmatrix}\vec i & \vec j & \vec k \\\frac{\partial  }{\partial x} & \frac{\partial  }{\partial y} & \frac{\partial  }{\partial z}\\ 3y & -xz& yz^2 \end{bmatrix} =(x+z^2,0,-z-3) \\ \text{Let }\vec r=(x,y,z)\text{ and }\cases{x=\rho\cos \theta\\ y=\rho \sin \theta\\ z=\rho^2/2} \Rightarrow \cases{\vec r_\rho=( \cos \theta, \sin \theta,\rho) \\\vec r_\theta =(-\rho \sin \theta, \rho \cos \theta, 0)} \\ \Rightarrow \vec r_\rho \times \vec r_\theta =(-\rho^2\cos \theta, -\rho^2\sin \theta, \rho) =(\rho^2\cos \theta, \rho^2\sin \theta, -\rho) \\ \Rightarrow \iint_S \text{curl }\vec F\cdot \vec n dS = \int_0^{2\pi} \int_0^2 (\rho \cos \theta+{\rho^4\over 4},0,-{\rho^2\over 2}-3) \cdot (\rho^2\cos \theta, \rho^2\sin \theta, -\rho)\,d\rho d\theta \\= \int_0^{2\pi} \int_0^2 (\rho^3\cos^2 \theta +{1\over 4}\rho^6\cos \theta+ {1\over 2}\rho^3+ 3\rho) \,d\rho d\theta = \int_0^{2\pi}  (4\cos^2 \theta+{128\over 28} \cos \theta+ 8)\,d\theta \\=\int_0^{2\pi}  (4\cos^2 \theta+8)\,d\theta =\int_0^{2\pi}  (2\cos 2\theta +10)\,d\theta =20\pi\\ \text{Finally, we have }\int_C \vec F\cdot d\vec r=20\pi =\iint_S \text{curl }\vec F\cdot \vec ndS$$

解答: $$\textbf{(a)}\;\sin^2 t={1\over 2}-{1\over 2}\cos (2t) \Rightarrow L\{\sin^2 t\} =L\left\{{1\over 2}-{1\over 2}\cos (2t) \right\} ={1\over 2s}-{1\over 2}\cdot {s\over s^2+4} \\  \Rightarrow L\left\{{\sin^2 t \over t}  \right\} =\int_s^\infty \left({1\over 2u}-  {u\over 2(u^2+4)} \right)\,du =\left. \left[ {1\over 2}\ln u-{1\over 4} \ln(u^2+4)\right] \right|_s^\infty \\ ={1\over 4}\left. \left[ \ln u^2- \ln(u^2+4)\right] \right|_s^\infty  ={1\over 4}\left. \left[ \ln {u^2\over u^2+4} \right] \right|_s^\infty ={1\over 4}\left(  \ln 1-\ln {s^2\over s^2+4}\right) \\ ={1\over 4} \ln{s^2+4\over s^2}. \bbox[red, 2pt]{QED}\\ \textbf{(b)}\; L\left\{{\sin^2 t\over t} \right\} =f(s)=\int_0^\infty {\sin^2 t\over t} e^{-st}\,dt ={1\over 4}\ln {s^2+4\over s^2} \\\quad \Rightarrow f(1)= \int_0^\infty {\sin^2 t\over t} e^{-t}\,dt ={1\over 4}\ln {1+4\over 1} =\bbox[red, 2pt]{{1\over 4} \ln 5}$$

解答: $$\textbf{(a)}\; f(x)=f(-x) \Rightarrow f(x) \text{ is even } \Rightarrow b_n=0\\ a_0={1\over 2\pi} \int_{-\pi/2}^{\pi/2} 1\,dx={1\over 2}\\ a_n={1\over \pi} \int_{-\pi/2}^{\pi/2} \cos(n x)\,dx ={2\over n\pi} \sin{n\pi \over 2},n=1,2,\dots \\ \Rightarrow f(x)=  a_0+ \sum_{n=1}^\infty a_n \cos(nx)   \Rightarrow\bbox[red, 2pt]{ f(x)={1\over 2}+ \sum_{n=1 }^\infty {2\over n\pi} \sin{n\pi \over 2} \cos(nx)} \\ \textbf{(b)}\; f(x)=  {1\over 2}+ \sum_{n=1 }^\infty {2\over n\pi} \sin{n\pi \over 2} \cos(nx) \\\qquad ={1\over 2}+ {2\over \pi}(\cos x- {1\over 3}\cos(3x)+{1\over 5}\cos (5x)-{1\over 7}\cos(7x)+ \cdots)\\ \Rightarrow f(0)=1={1\over 2} +{2\over \pi}(1-{1\over 3} +{1\over 5} -{1\over 7} + \cdots) \Rightarrow 1-{1\over 3} +{1\over 5} -{1\over 7} + \cdots=\bbox[red, 2pt]{\pi \over 4}$$

解答: $$\phi(x,t)=X(x)T(t) \Rightarrow XT''=c^2X''T \Rightarrow \text{BC:} \cases{\phi_x(0,t) = X'(0)T(t)=0\\ \phi_x(\ell,t)=X'(\ell)T(t)=0} \Rightarrow \cases{X'(0)=0\\ X'(\ell)=0} \\ XT''=c^2X''T \Rightarrow {T''\over c^2T}={X''\over X} =\lambda\\ \textbf{Case I: }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow X'(0)=X'(\ell)=c_1 =0 \Rightarrow X=c_2\\\qquad \text{Taking }c_2=1,\text { we get }X=1\\ \textbf{Case II: }\lambda=\rho^2 \gt 0 \Rightarrow  \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x}+ c_2e^{-\rho x} \Rightarrow X'=c_1\rho e^{\rho x}-c_2\rho  e^{-\rho x} \\ \qquad \cases{X'(0)=\rho(c_1-c_2)=0\\ X'(\ell)=c_1\rho e^{\rho \ell}-c_2 \rho e^{-\rho \ell} =0} \Rightarrow c_1=c_2=0 \Rightarrow  X=0\\ \textbf{Cases 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'=-c_1\rho \sin(\rho x)+ c_2\rho \cos(\rho x) \Rightarrow X'(0)=c_2\rho =0 \Rightarrow c_2=0 \\\qquad \Rightarrow X'(\ell)= -c_1\rho \sin(\rho \ell)=0 \Rightarrow \sin(\rho \ell)=0 \Rightarrow \rho \ell = n\pi \Rightarrow \rho ={n\pi\over \ell}\\\qquad  \Rightarrow X=c_1 \cos({n \pi x\over \ell}) \Rightarrow X_n=\cos({n \pi x\over \ell}),n=1,2,\dots\\ \qquad \Rightarrow T''=\lambda c^2 T \Rightarrow T''+{n^2\pi^2 \over \ell^2}c^2 T=0 \Rightarrow T_n =c_1 \cos{nc\pi \over \ell}t +c_2\sin{nc\pi \over \ell}t,n=1,2,\dots \\ \Rightarrow \phi(x,t)=a_0+ \sum_{n=1}^\infty (A_n \cos{nc \pi t\over \ell} +B_n \sin{nc \pi t\over \ell}) \cos{n\pi x\over \ell} \\ \Rightarrow \phi(x,0)= a_0+ \sum_{n=1}^\infty A_n \cos{n\pi x\over \ell} =e^{-x} \Rightarrow a_0= {1 \over \ell}\int_0^\ell e^{-x}\,dx ={1\over \ell}(1-e^{-\ell}) \\ \Rightarrow A_n = {2\over \ell} \int_0^\ell e^{-x} \cos{n\pi x \over \ell}\,dx  ={2\ell (1-e^{-\ell}(-1)^n \over n^2\pi^2 +\ell^2}\\ \phi_t(x,0)=0 \Rightarrow B_n=0 \Rightarrow \bbox[red, 2pt]{\phi(x,t)={1\over \ell}(1-e^{-\ell}) + \sum_{n=1}^\infty {2\ell (1-e^{-\ell}(-1)^n \over n^2\pi^2 +\ell^2} \cos{nc\pi t\over \ell} \cos{n\pi x\over \ell}}$$

解答: $$z^4+1=0 \Rightarrow z^4=e^{i(\pi+2k\pi)} \Rightarrow z=e^{i(\pi +2k\pi) \over 4},k=0,1,2,3 \\ \Rightarrow z_0= e^{i\pi/4} ={\sqrt 2\over 2}+i{\sqrt 2\over 2},z_1={\sqrt 2\over 2}-i{\sqrt 2\over 2} \\ \int_{-\infty}^\infty {1\over x^4+1}\,dx = \oint_C {1\over z^4+1}\,dz = 2\pi i \left(\left. {1\over 4z^3}\right|_{z=z_0} +\left.{1\over 4z^3}\right|_{z=z_1} \right) =\bbox[red, 2pt]{\sqrt 2\pi \over  2}$$

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