115年台大碩士班-工程數學G詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 工程數學(G)

解答:$$\textbf{(a) } \det(A) =-\cos^2(2\alpha)-\sin^2(2\alpha) = \bbox[red, 2pt]{-1} \\\textbf{(b) } A^{-1} ={1\over \det (A)} \begin{bmatrix}-\cos(2\alpha) &-\sin(2\alpha) \\ -\sin(2\alpha)& \cos(2\alpha) \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}\cos(2\alpha) &\sin(2\alpha) \\ \sin(2\alpha)& -\cos(2\alpha) \end{bmatrix}} \\\textbf{(c) } A\mathbf x = \begin{bmatrix}\cos(2\alpha) &\sin(2 \alpha) \\ \sin(2\alpha)& -\cos(2\alpha) \end{bmatrix} \begin{bmatrix} \cos \theta\\ \sin \theta \end{bmatrix} = \begin{bmatrix} \cos(2\alpha) \cos \theta+ \sin(2 \alpha) \sin \theta \\ \sin(2\alpha)\cos \theta-\cos(2\alpha)\sin \theta \end{bmatrix} = \begin{bmatrix}\cos(2\alpha-\theta) \\ \sin(2\alpha-\theta) \end{bmatrix} \\\qquad \Rightarrow \bbox[red, 2pt]{\beta=2\alpha-\theta} \\\textbf{(d) } \det(A-\lambda I) = \begin{vmatrix} \cos(2\alpha)-\lambda& \sin(2 \alpha) \\ \sin(2\alpha)&-\cos(2\alpha)-\lambda \end{vmatrix} = \lambda^2-(\cos^2(2\alpha)+\sin^2(2\alpha)) =\lambda^2-1=0 \\ \Rightarrow \lambda= \bbox[red, 2pt]{\pm 1} \\ \textbf{(e) }\lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} \cos(2\alpha)-1& \sin(2\alpha) \\ \sin(2\alpha)&-\cos(2 \alpha) -1\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1+{\sin(2\alpha) \over \cos(2\alpha)-1}x_2=0 \\ \qquad \Rightarrow v= x_2 \begin{bmatrix}-{\sin(2\alpha) \over \cos(2\alpha)-1} \\1 \end{bmatrix} , \text{ Choosing }x_2=1, \text{ we get }v_1=  \bbox[red, 2pt]{\begin{bmatrix}-{\sin(2\alpha) \over \cos(2\alpha)-1} \\1 \end{bmatrix}} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix} \cos(2\alpha)+1& \sin(2 \alpha) \\ \sin(2\alpha)&-\cos(2\alpha)+1\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1+{\sin(2\alpha) \over \cos(2\alpha)+1}x_2=0 \\ \qquad \Rightarrow v= x_2 \begin{bmatrix} {-\sin(2\alpha) \over \cos(2\alpha)+1} \\1\end{bmatrix} , \text{ Choosing }x_2=1, \text{ we get }v_2= \bbox[red, 2pt]{\begin{bmatrix} {-\sin(2\alpha) \over \cos(2\alpha)+1} \\1\end{bmatrix}}$$

解答:$$\textbf{(a) }y''+2y'+3y=0 \Rightarrow r^2+2r+3=0 \Rightarrow r=-1\pm \sqrt 2 i \Rightarrow y=e^{-x}(c_1 \cos \sqrt 2 x+c_2\sin \sqrt 2x) \\\quad \Rightarrow y'=e^{-x}((\sqrt 2c_2-c_1)\cos \sqrt 2 x +(-\sqrt 2c_1-c_2)\sin \sqrt 2x)\\ \Rightarrow \cases{y(0)=c_1=2\\ y'(0)=\sqrt 2 c_2-c_1=-3 } \Rightarrow \cases{c_1=2\\ c_2=-\sqrt 2/2} \Rightarrow \bbox[red, 2pt]{y=e^{-x} \left( 2\cos \sqrt 2x-{\sqrt2 \over   2}\sin \sqrt 2x  \right)} \\\textbf{(b) }y''-12y'+36y=0 \Rightarrow r^2-12r+36=0 \Rightarrow (r-6)^2=0 \Rightarrow r=6\\ \quad \Rightarrow \bbox[red, 2pt]{y=c_1e^{6x}+ c_2 xe^{6x}} \\\textbf{(c) }y''+4y=0 \Rightarrow r^2+4=0 \Rightarrow r=\pm 2i \Rightarrow y_h= c_1\cos 2x+ c_2\sin 2x\\ \quad y_p=Ae^x \Rightarrow y_p'=y_p''=Ae^x \Rightarrow y_p''+4y_p=5Ae^x=e^x \Rightarrow A={1\over 5} \Rightarrow y_p={1\over 5}e^x\\ \quad \Rightarrow y= y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos 2x+ c_2\sin 2+ {1\over 5}e^x}$$

解答:$$\textbf{(a) } \bbox[red, 2pt]{\text{Second-order}} \\ \textbf{(b) } u(r,\theta) =R(r)\Theta(\theta) \Rightarrow {1\over r} {\partial \over \partial r} \left( rR'\Theta \right)+{1\over r} R\Theta''= 0 \Rightarrow {r^2R''\over R} =-{\Theta''\over \Theta} = \lambda^2 \\ \Rightarrow \cases{\Theta''+\lambda^2 \Theta=0\\ r^2R''+rR'-\lambda^2 R=0} \\ \text{Solving for }\Theta: \Theta''+ \lambda^2 \Theta=0 \Rightarrow \Theta= A\cos(\lambda \theta)+ B\sin(\lambda \theta ) \Rightarrow \Theta'=-A\lambda \sin(\lambda \theta)+ B\lambda \cos(\lambda\theta) \\ \qquad B.C.:\cases{u(r,-\pi) =u(r,\pi) \\ {\partial u\over \partial \theta}(r,-\pi) = {\partial u\over \partial \theta} (r,\pi)} \Rightarrow \sin(\lambda \pi)=0 \Rightarrow \lambda =n \Rightarrow \Theta_n(\theta) =A_n \cos(n\theta)+ B_n \sin(n\theta), n=0,1,2,\dots \\ \text{Solving for }R: r^2R''+rR'-n^2R=0 \\ \textbf{Case I }n=0 \Rightarrow r^2R''+rR'=0 \Rightarrow R_0=C_0+D_0\ln r \\\textbf{Case II }n\gt 0 \Rightarrow R=r^k \Rightarrow (k(k-1)+k-n^2)r^k=0 \Rightarrow k^2-n^2=0\Rightarrow k=\pm n \Rightarrow R_n=C_nr^n+D_nr^{-n} \\ B.C.: \lim_{r\to 0}u(r,\theta) \lt \infty \Rightarrow \cases{D_0=0 \\ D_n=0} \Rightarrow \cases{R_0=C_0\\ R_n= r^n} \\ \Rightarrow u(r,\theta) ={a_0\over 2}+ \sum_{n=1}^\infty r^n(a_n\cos n\theta+ b_n\sin n\theta) \\\Rightarrow u({5\over 3},\theta) =\ln 2+ 4\cos 3\theta ={a_0\over 2}+ \sum_{n=1}^\infty (5/3)^n(a_n\cos n\theta+ b_n\sin n\theta) \\ \Rightarrow \cases{a_0/2= \ln 2\\ n=3 \Rightarrow (5/3)^3 a_3\cos 3\theta= 4\cos 3\theta \Rightarrow a_3=108/125\\ b_n=0} \Rightarrow \bbox[red, 2pt]{u(r,\theta)= \ln 2+ {108\over 125}r^3 \cos 3\theta}$$

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解題僅供參考，碩士班歷年試題及詳解