113年成大環工碩士班-工程數學詳解

 國立成功大學113學年度碩士班招生考試

系所：環境工程學系
科目:工程數學

解答: $$\textbf{A.}\;\text{Case I }\lambda=0 \Rightarrow y''=0 \Rightarrow y(x) =c_1+c_2x \Rightarrow \cases{y(0)=0= c_1 \\ y(3)=0=c_1+3c_2} \Rightarrow c_1=c_2=0\\\qquad \Rightarrow y=0 \Rightarrow \text{ no eigenvalue} \\\text{Case II }\lambda \gt 0 \Rightarrow r^2+\lambda=0 \Rightarrow r=\pm \sqrt{\lambda }i \Rightarrow y=c_1\cos(\sqrt{\lambda }x)+ c_2\sin(\sqrt \lambda x) \\ \qquad \Rightarrow \cases{y(0)=c_1=0\\ y(3)=c_1\cos(3\sqrt \lambda)+ c_2\sin (3\sqrt{\lambda}) =0} \Rightarrow \sin(3\sqrt \lambda)=0 \Rightarrow 3\sqrt \lambda=n\pi  \\ \qquad \Rightarrow \lambda_n={n^2\pi^2\over 9} \Rightarrow y_n(x)=\sin({n\pi\over 3} x),n=1,2,\dots \\ \text{Case III }\lambda\lt 0 \Rightarrow r=\pm \sqrt{-\lambda } \Rightarrow y(x)= c_1\cosh(\sqrt{-\lambda }x)+ c_2\sinh(\sqrt{-\lambda }x) \\ \qquad \Rightarrow \cases{y(0)=c_1=0\\ y(3)=c_2\sinh(\sqrt{-\lambda }x)=0} \Rightarrow c_1=c_2=0 \Rightarrow y=0 \Rightarrow \text{ no eigenvalue}\\ \text{In summary, we have eigenvalues: }\bbox[red, 2pt]{\lambda_n={n^2\pi^2\over 9}} \text{ and eigenfunctions: } \bbox[red, 2pt]{y_n(x)=\sin({n\pi\over 3}x)} ,n=1,2,3,\cdots \\\textbf{B.}\;\text{Case I }\lambda=0 \Rightarrow y''=0 \Rightarrow y(x) =c_1+c_2x \Rightarrow y'(x)=c_2 \\\qquad \Rightarrow \cases{y(0)=0= c_1 \\ y'(\pi)=0=c_1+c_2 \pi} \Rightarrow c_1=c_2=0  \Rightarrow y=0 \Rightarrow \text{ no eigenvalue} \\\text{Case II }\lambda \gt 0 \Rightarrow r^2+\lambda=0 \Rightarrow r=\pm \sqrt{\lambda }i \Rightarrow y=c_1\cos(\sqrt{\lambda }x)+ c_2\sin(\sqrt \lambda x) \\ \qquad \Rightarrow  y'=-c_1\sqrt{\lambda} \sin(\sqrt{\lambda }x)+ c_2 \sqrt{\lambda} \cos(\sqrt{\lambda }x) \\\cases{y(0)=c_1=0\\ y'(\pi)=  c_2\sqrt{\lambda}\cos (\sqrt{\lambda}\pi) =0} \Rightarrow \cos(\sqrt \lambda \pi)=0 \Rightarrow \sqrt \lambda \pi={2n-1\over 2}\pi  \\ \qquad \Rightarrow \lambda_n={(2n-1)^2 \over 4} \Rightarrow y_n(x)=\sin({2n-1\over 2} x),n=1,2,\dots \\ \text{Case III }\lambda\lt 0 \Rightarrow r=\pm \sqrt{-\lambda } \Rightarrow y(x)= c_1\cosh(\sqrt{-\lambda }x)+ c_2\sinh(\sqrt{-\lambda }x) \\ \qquad \Rightarrow y'=c_1 \sqrt{-\lambda} \sinh (\sqrt{-\lambda }x) +c_2 \sqrt{-\lambda} \cosh(\sqrt{-\lambda}x) \\ \qquad \Rightarrow \cases{y(0)=c_1=0\\ y'(\pi)=c_2 \sqrt{-\lambda}\cosh(\sqrt{-\lambda }\pi)=0} \Rightarrow c_1=c_2=0 \Rightarrow y=0 \Rightarrow \text{ no eigenvalue}\\ \text{In summary, we have eigenvalues: }\bbox[red, 2pt]{\lambda_n={(2n-1)^2 \over 4}} \text{ and eigenfunctions: }\\\qquad \bbox[red, 2pt]{y_n(x)=\sin({2n-1\over 2}x)} ,n=1,2,3,\cdots$$

解答: $$\cases{x'=2x-7y\\ y'=5x+10y+4z\\ z'=5y+2z} \Rightarrow \begin{bmatrix}x' \\y'\\ z' \end{bmatrix} =\begin{bmatrix}2 & -7& 0 \\5 & 10& 4 \\ 0 &5& 2\end{bmatrix}\begin{bmatrix}x \\y\\ z \end{bmatrix} \equiv X'=AX\\ \det(A-\lambda I)=0 \Rightarrow \lambda=2,5,7\\ \lambda_1=2 \Rightarrow \text{ eigenvector }v_1=\begin{pmatrix}-4 \\0\\5 \end{pmatrix},\lambda_2=5 \Rightarrow \text{ eigenvector }v_2=\begin{pmatrix}-7 \\3\\5 \end{pmatrix},\\\lambda_3=7 \Rightarrow \text{ eigenvector }v_3=\begin{pmatrix}-7 \\5 \\5 \end{pmatrix} \Rightarrow X=c_1e^{2t}v_1 + c_2e^{5t}v_2 + c_3e^{7t}v_3 \\ \Rightarrow \bbox[red, 2pt]{\cases{x(t)=-4c_1e^{2t} -7c_2e^{5t} -7c_3e^{7t} \\ y(t)= 3c_2e^{5t}+5c_3e^{7t} \\z(t)= 5c_1e^{2t} +5c_2e^{5t} +5c_3e^{7t} }}$$

解答: $$y'' +4y'+13y=0 \Rightarrow \lambda^2+4\lambda+13=0 \Rightarrow \lambda={-4\pm 6i\over 2} =-2\pm 3i \\ \Rightarrow \bbox[red, 2pt]{y=e^{-2x}(A\cos(3x)+ B\sin(3x))}$$

解答: $$y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=1,2\\ \Rightarrow y_h= c_1e^x+ c_2e^{2x}\\ y_p=A\cos(x)+B\sin (x) \Rightarrow y_p'= -A\sin(x)+B\cos(x) \Rightarrow y_p''= -A\cos(x)-B\sin(x)\\ \Rightarrow y_p''-3y_p'+2y_p = (A-3B)\cos(x)+(3A+B)\sin x=10\sin(x)\\ \Rightarrow \cases{A-3B=0\\ 3A+B=10} \Rightarrow \cases{A=3\\ B=1} \Rightarrow y_p=3\cos(x)+\sin(x) \Rightarrow y=y_h+y_p \\\ \Rightarrow\bbox[red, 2pt]{ y=c_1e^x+ c_2e^{2x}+ 3\cos(x)+ \sin (x)}$$

解答: $${dy\over dx}={4x+6y\over 10y-6x} \Rightarrow (6x-10y)dy+( 4x+6y)dx =0 \Rightarrow \frac{\partial  }{\partial x}(6x-10y)=6=\frac{\partial  }{\partial y} ( 4x+6y) \\ \Rightarrow \text{it is exact } \Rightarrow \Phi(x,y)=\int (6x-10y)dy =\int( 4x+6y)dx  \\ \Rightarrow \Phi(x,y)=6xy-5y^2+ \rho(x)=2x^2+6xy+ \phi(x) \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{2x^2+6xy-5y^2+c_1=0}$$

解答: $$\lambda^2+7\lambda=0 \Rightarrow \lambda=0,-7 \Rightarrow y=c_1+ c_2e^{-7x} \Rightarrow y'=-7c_2e^{-7x} \\ \Rightarrow \cases{y(0)=c_1+c_2=1\\ y'(0)= -7c_2=7} \Rightarrow \cases{c_1 =2\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y=2-e^{-7x}}$$

解答: $$a_0={1\over 2\pi}\int_{-\pi}^\pi f(x)\,dx = {1\over 2\pi}\int_{0}^\pi 3\,dx ={3\over 2} \\ a_n= {1\over \pi} \int_0^\pi 3\cos(nx)\,dx=0\\ b_n={1\over \pi} \int_0^\pi 3\sin(nx)\,dx ={3\over n\pi}(1-(-1)^n ) \\ \Rightarrow f(x)=\bbox[red, 2pt]{{3\over 2} +{3\over \pi}\sum_{n=1}^\infty {1\over n}(1-(-1)^n )}$$

解答: $$\textbf{A.} \text{Let }u(x,y)= X(x)Y(y), \text{ then }u_{xx}+u_{yy}=0 \Rightarrow X''Y+ XY''=0 \\ \Rightarrow {X''\over X}=-{Y''\over Y}=\lambda \Rightarrow \cases{X''=\lambda X\\ Y''=-\lambda Y} \\ \text{boundary conditions: (B.C.)} \cases{u_x(0,y)=0\\ u_x(1,y)=0 \\ u(x,0)=0} \Rightarrow \cases{X'(0)=0\\ X'(1)=0 \\ Y(0)=0} \\ \text{Case I }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+ c_2 \Rightarrow X'=c_1 \Rightarrow X'(0)=0 \Rightarrow c_1=0\\ \qquad \Rightarrow X=c_2\\ \text{Case II }\lambda \gt 0 \Rightarrow \lambda=\rho^2\text{ for some }\rho \gt 0 \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1e^{\rho x} +c_2 e^{-\rho x} \\\qquad \Rightarrow X'=c_1\rho e^{\rho x}-c_2 \rho e^{-\rho x} \Rightarrow B.C. \cases{\rho(c_1-c_2)=0\\ \rho(c_1e^{2}-c_2)=0} \Rightarrow c_1=c_2=0\\ \qquad \Rightarrow X=0\\ \text{Case III } \lambda\lt 0 \Rightarrow \lambda=-\rho^2 \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 B.C. \cases{c_2\rho=0\\ -c_1\rho\sin(\rho)+c_2 \rho\cos(\rho)=0} \\ \qquad \Rightarrow c_2=0 \Rightarrow \sin(\rho)=0 \Rightarrow \rho=n\pi \Rightarrow X=c_1\cos(n\pi x)\\ \text{Now we need to find the corresponding }Y.\\ \text{Case I }\lambda=0 \Rightarrow Y''=0 \Rightarrow Y=c_1y+c_2 \Rightarrow B.C. y(0)=c_2=0 \Rightarrow y=c_1y \\\qquad \Rightarrow u(x,y)=c_1y\cdot c_2 =C_0y \\\text{Case III }\lambda\lt 0 \Rightarrow \lambda=-\rho^2 =-n^2\pi^2 \Rightarrow Y''=n^2\pi^2 Y \Rightarrow Y=c_1 \cosh(n\pi y)+ c_2\sinh(n\pi y) \\\qquad \Rightarrow B.C. y(0)=c_1=0 \Rightarrow Y=c_2\sinh(n\pi y) \\\qquad \Rightarrow u(x,y)=  c_1c_2 \cos(n\pi x) \sinh(n\pi y) =C_n\cos(n\pi x) \sinh(n\pi y) \\ \text{At last, we have }{u(x,y)=C_0y+ \sum_{n=1}^\infty C_n \cos(n\pi x)\sinh(n\pi y)} \\ \Rightarrow u(x,1)=5 \Rightarrow 5=C_0+\sum_{n=1}^\infty C_n \cos(n\pi x)\sinh(n\pi ) \Rightarrow C_0= \int_0^1 5\,dx=5 \\ \Rightarrow C_n \sinh(n\pi)= 2\int_0^1 5\cos(n\pi x)\,dx=0 \Rightarrow C_n=0 \Rightarrow \bbox[red, 2pt]{u(x,y)=5y}$$

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