114年中興土木碩士班-工程數學詳解

 國立中興大學114學年度碩士班考試入學招生考試

科目：工程數學
系所：土木工程學系甲組
本科目依簡章規定「可以」使用計算機

解答:$$ay''''=1-x \Rightarrow y''''={1\over a}(1-x) \Rightarrow y'''=\int {1\over a}(1-x)\,dx ={1\over a}(x-{1\over 2}x^2)+c_1 \\ \Rightarrow y'''(1)={1\over 2a}+c_1=0 \Rightarrow c_1=-{1\over 2a} \Rightarrow y'''={1\over a}(-{1\over 2}x^2+x-{1\over 2}) \\ \Rightarrow y''= \int {1\over a}(-{1\over 2}x^2+x-{1\over 2})\,dx  ={1\over a} \left( -{1\over 6}x^3+{1\over 2}x^2-{1\over 2}x \right) +c_2 \\ \Rightarrow y''(0)=c_2=0 \Rightarrow y''={1\over a} \left( -{1\over 6}x^3+{1\over 2}x^2-{1\over 2}x \right) \Rightarrow y'= \int {1\over a} \left( -{1\over 6}x^3+{1\over 2}x^2-{1\over 2}x \right) \,dx \\={1\over a} \left( -{1\over 24}x^4+{1\over 6}x^3-{1\over 4}x^2 \right) +c_3 \Rightarrow y'(1)={1\over a} \left( -{1\over 8}\right)+c_3=0 \Rightarrow c_3={1\over 8a} \\ \Rightarrow y'={1\over a} \left( -{1\over 24}x^4+{1\over 6}x^3-{1\over 4}x^2 +{1\over 8}\right) \Rightarrow y= \int {1\over a} \left( -{1\over 24}x^4+{1\over 6}x^3-{1\over 4}x^2 +{1\over 8}\right) \,dx  \\={1\over a} \left( -{1\over 120}x^5+{1\over 24}x^4-{1\over 12}x^3 +{1\over 8}x\right) +c_4 \Rightarrow y(0)=c_4=0 \\ \Rightarrow \bbox[red, 2pt]{y= {1\over a} \left( -{1\over 120}x^5+{1\over 24}x^4-{1\over 12}x^3 +{1\over 8}x\right)}$$

解答:$$y'=(x^2+9)(y^2-9) \Rightarrow \int{1\over y^2-9}\,dy = \int (x^2+9)\,dx \Rightarrow  {1\over 6}\ln {y-3\over y+3} ={1\over 3}x^3+9x+c_1 \\ \Rightarrow \ln {y-3\over y+3} =2x^3+54x +c_2 \Rightarrow {y-3\over y+3}=1-{6\over y+3} =c_3e^{2x^3+54x} \\\Rightarrow y={6\over 1-c_3e^{2x^3+54x}}-3 \Rightarrow \bbox[red, 2pt]{y= 3\cdot {1+c_3e^{2x^3+54x} \over 1-c_3e^{2x^3+54x}}}$$

解答:$$L\{y''\} +L\{y\} =L\{H(t-2) \sin(t-2)\} \Rightarrow s^2Y(s)+Y(s)= {e^{-2s} \over s^2+1} \Rightarrow Y(s) = {e^{-2s} \over (s^2+1)^2} \\ \Rightarrow y(t)= L^{-1}\left\{ {e^{-2s} \over (s^2+1)^2} \right\} =H(t-2) L^{-1}\left\{ {e^{-2s} \over (s^2+1)^2} \right\}(t-2) \\ \Rightarrow \bbox[red, 2pt]{y(t)={1\over 2} H(t-2)\left(\sin(t-2)-(t-2)\cos(t-2) \right)}$$

解答:$$\det(A-\lambda I) = \begin{vmatrix} 4-\lambda & -2 & 0 \\3 & -3-\lambda & -1 \\0 & 0 & 0-\lambda\end{vmatrix} = -\lambda(\lambda+2)(\lambda-3)=0 \Rightarrow \lambda=0,-2,3 \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow \begin{bmatrix} 4 & -2 & 0 \\3 & -3 & -1 \\0 & 0 & 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{3x_1+x_3=0\\ 3x_2+2x_3=0} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} -1/3\\ -2/3\\ 1\end{pmatrix}, \text{ choose }v_1= \begin{pmatrix} -1/3\\ -2/3\\ 1\end{pmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow \begin{bmatrix} 6 & -2 & 0 \\3 & -1 & -1 \\0 & 0 & 2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{3x_1=x_2\\ x_3=0} \\\qquad \Rightarrow v= x_2 \begin{pmatrix} 1/3\\ 1\\ 0\end{pmatrix}, \text{ choose }v_2= \begin{pmatrix} 1/3\\ 1\\ 0 \end{pmatrix} \\ \lambda_3= 3\Rightarrow (A-\lambda_3 I) v=0 \Rightarrow \begin{bmatrix} 1 & -2 & 0 \\3 & -6 & -1 \\0 & 0 & -3\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=2x_2\\ x_3=0} \\\qquad \Rightarrow v= x_2 \begin{pmatrix} 2\\ 1\\ 0\end{pmatrix}, \text{ choose }v_3= \begin{pmatrix} 2\\ 1\\ 0 \end{pmatrix} \\ \Rightarrow \textbf{(a) } \bbox[red, 2pt]{\text{ The eigenvalues of }A= 0,-2,3} \\ \textbf{(b) } \bbox[red, 2pt]{\text{The corresponding eigenvectors: }  \begin{pmatrix} -1/3\\ -2/3\\ 1\end{pmatrix},\begin{pmatrix} 1/3\\ 1\\ 0 \end{pmatrix},\begin{pmatrix} 2\\ 1\\ 0 \end{pmatrix}}$$

解答:$$u(x,t) =X(x)T(t) \Rightarrow \text{ BC. }\cases{u(0,t)= X(0)T(t)=0 \\ u(L,t)= X(L)T(t)=0} \Rightarrow \cases{ X(0)=0\\ X(L)=0} \\ {\partial u\over\partial t} = \alpha{\partial^2 u\over \partial x^2} \Rightarrow XT'=\alpha X''T \Rightarrow {T'\over \alpha T} ={X''\over X} =k \Rightarrow \cases{X''-kX=0\\ T'-k \alpha T=0} \\ \textbf{Case I }k=0 \Rightarrow X''=0 \Rightarrow X=c_1 x+c_2 \Rightarrow \cases{X(0)=c_2=0 \\ X(L) =c_1L+ c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow X=0  \\ \textbf{Case II }k=\mu^2 \gt 0 \Rightarrow X''-\mu^2X=0 \Rightarrow X=c_1 e^{\mu x}+ c_2e^{-\mu x} \Rightarrow \cases{X(0)= c_1+c_2=0\\ X(L) =c_1e^{\mu L}+c_2 e^{-\mu L}=0} \\\qquad \Rightarrow c_2=-c_1 \Rightarrow c_1e^{\mu L}-c_1e^{-\mu L} =0 \Rightarrow c_1(e^{2\mu L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2= 0 \Rightarrow X=0 \\\textbf{Case III }k=-\mu^2 \lt 0 \Rightarrow X''+\mu^2 X=0 \Rightarrow X= c_1 \cos \mu x +c_2\sin \mu x \Rightarrow X(0)=c_1= 0 \\\qquad \Rightarrow  X(L)= c_2\sin \mu L=0 \Rightarrow \mu L= n\pi \Rightarrow \mu= {n\pi \over L} \\\qquad \Rightarrow X_n= \cos{n\pi x\over L}, n\in \mathbb N \text{ (choosing c_1=1)} \\\Rightarrow T'+k \mu^2 T=0 \Rightarrow T_n= e^{-k\mu^2 t} =e^{-k{n^2\pi^2 t/L^2}} \Rightarrow u_n=X_nT_n =e^{-k{n^2\pi^2 t/L^2}} \cos{n\pi x\over L}\\ \Rightarrow u(x,t) =\sum_{n=1}^\infty a_n e^{-k{n^2\pi^2t /L^2}} \cos{n\pi x\over L} \Rightarrow u(x,0)= u_0 \sin({\pi x\over L}) = \sum_{n=1}^\infty a_n \cos{n\pi x\over L} \\ \Rightarrow a_n= {2\over L}\int_0^L u_0\sin{\pi x\over L} \cos {n\pi x\over L}\,dx ={2u_0 \over L} \cdot {L(1+(-1)^n)\over \pi(1-n^2)} ={2u_0(1+(-1)^n) \over \pi(1-n^2)} \\ \Rightarrow \bbox[red, 2pt]{u(x,t) ={2u_0\over \pi}\sum_{n=1}^\infty { 1+(-1)^n \over 1-n^2} e^{-k{n^2\pi^2t /L^2}} \cos{n\pi x\over L}}$$

解答:$${\partial \over \partial t}u(x,t)+ c{\partial \over \partial x}u(x,t)=0 \Rightarrow \left( {\partial u\over \partial t}, {\partial u\over \partial x}\right) \cdot (1, c) =0 \Rightarrow \left( {\partial u\over \partial t}, {\partial u\over \partial x}\right) \bot (1,c) \\ \Rightarrow {d x\over d t}=c \Rightarrow x =ct+x_0 \Rightarrow x_0=x-ct \Rightarrow u(x,t) =f(x-ct) \Rightarrow u(x,t)= \begin{cases}1& 0\le x-ct\le L\\ 0& \text{otherwise} \end{cases} \\ \Rightarrow  \bbox[red, 2pt]{u(x,t)= \begin{cases}1& ct\le x \le L+ct\\ 0& \text{otherwise} \end{cases} }$$

解答:$$\cases{x(t)=t\\ y(t)=t/2 \\z(t)=t/3}  \Rightarrow \cases{x'(t)=1\\ y'(t)=1/2\\ z'(t)=1/3}, 0\le t\le 1 \\ \Rightarrow \int_C \vec F\cdot d\vec r = \int_0^1  \left( t^2, {t^2\over 4}, {t^2\over 9}\right) \cdot \left(1,{1\over 2},{1\over 3} \right)\,dt = \int_0^1   {251\over 216}t^2\,dt =\bbox[red, 2pt] {251\over 648}$$

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