國立臺灣科技大學111學年度碩士班招生考試
系所組別:機械工科甲乙丙丁組
科目:工程數學
解答:$$\textbf{(a)}\;y''-16y=0 \Rightarrow y_h=c_1e^{4x}+ c_2e^{-4x}\\ y_p=Axe^{4x} \Rightarrow y_p'=Ae^{4x}+4Axe^{4x} \Rightarrow y_p''=8Ae^{4x} + 16Axe^{4x} \\ \Rightarrow y_p''-16y_p=8Ae^{4x} =2e^{4x} \Rightarrow A={1\over 4}\\ y=y_h+y_p\Rightarrow \bbox[red, 2pt]{y =c_1e^{4x}+ c_2e^{-4x} +{1\over 4}xe^{4x}} \\\textbf{(b)}\;y''+y=0 \Rightarrow y_h= c_1\cos x+c_2\sin x\\ \text{Let }\cases{y_1=\cos x\\ y_2=\sin x} \Rightarrow W=\begin{vmatrix} y_1& y_2 \\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} \cos x& \sin x \\-\sin x& \cos x \end{vmatrix} =1 \\ \Rightarrow y_p =-\cos x\int\sin x(4x+10\sin x)\,dx +\sin x \int \cos x(4x+10\sin x)\,dx \\= -\cos x(5x+4\sin x-4x\cos x-5\sin x\cos x) +\sin x(4x \sin x- 5\cos^2 x+4 \cos x) \\=-5x\cos x+4x \Rightarrow y=y_h+y_p = c_1\cos x+c_2\sin x-5x\cos x+4x\\ \Rightarrow y'=-c_1\sin x +c_2 \cos x-5\cos x+5x\sin x+4 \\ \Rightarrow \cases{y(\pi)= -c_1+9\pi=0\\ y'(\pi)= -c_2+9=2} \Rightarrow \cases{c_1=9\pi \\ c_2=7} \Rightarrow \bbox[red, 2pt]{y =9\pi \cos x+7\sin x-5x\cos x+4x}$$
解答:$$\vec a_1=\vec v_1 =\begin{pmatrix} 1\\ 1\\1\\-1\end{pmatrix} \Rightarrow \vec e_1={\vec a_1 \over \Vert \vec a_1\Vert} =\left(\begin{matrix}\frac{1}{2} \\\frac{1}{2} \\\frac{1}{2} \\\frac{-1}{2} \end{matrix} \right) \\ \vec a_2 =\vec v_2-(\vec v_2\cdot \vec e_1)\vec e_1 =\left(\begin{matrix} \frac{9}{4} \\\frac{-3}{4} \\\frac{-3}{4} \\\frac{3}{4} \end{matrix} \right) \Rightarrow \vec e_2={\vec a_2\over \Vert \vec a_2\Vert }=\left(\begin{matrix} \frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{6} \\ \frac{-\sqrt{3}}{6} \\ \frac{\sqrt{3}}{6} \end{matrix} \right) \\ \vec a_3= \vec v_3-(\vec v_3\cdot \vec e_1)\vec e_1 -(\vec v_3\cdot \vec e_2)\vec e_2 =\vec v_3- \vec e_1 + \sqrt 3\vec e_2 =\left(\begin{matrix} 0 \\1 \\1 \\2 \end{matrix} \right)\\ \Rightarrow \vec e_3={\vec a_1\over \Vert \vec a_3\Vert} =\left( \begin{matrix} 0 \\ \frac{\sqrt{6}}{6} \\ \frac{\sqrt{6}}{6} \\ \frac{\sqrt{6}}{3}
\end{matrix}\right) \\ \Rightarrow \text{orthonormal vector set }=\bbox[red, 2pt]{\left\{ \left(\begin{matrix}\frac{1}{2} \\\frac{1}{2} \\\frac{1}{2} \\\frac{-1}{2} \end{matrix} \right), \left(\begin{matrix} \frac{\sqrt{3}}{2} \\ \frac{-\sqrt{3}}{6} \\ \frac{-\sqrt{3}}{6} \\ \frac{\sqrt{3}}{6} \end{matrix} \right) ,\left( \begin{matrix} 0 \\ \frac{\sqrt{6}}{6} \\ \frac{\sqrt{6}}{6} \\ \frac{\sqrt{6}}{3} \end{matrix}\right) \right\}}$$
解答:$$L^{-1}\left\{{s+3\over (s+2)(s^2+2s+2} \right\}=L^{-1} \left\{{1\over 2(s+2)} +{-s+2\over 2(s^2+2s+2)} \right\}\\=L^{-1}\left\{ {1\over 2(s+2)}-{1\over 2} \cdot {s+1\over (s+1)^2+1} +{3\over 2}\cdot {1\over (s+1)^2+1}\right\}\\ =\bbox[red, 2pt]{{1\over 2}e^{-2t}-{1\over 2}e^{-t} \cos t +{3\over 2}e^{-t} \sin t }$$
解答:$$A=\begin{bmatrix} 2& 0 & 1\\ 2& -1 & 0\\ -1& 0 & 0\end{bmatrix} \xrightarrow{ 2R_3+R_1\to R_1,2R_3+R_2 \to R_2} \begin{bmatrix} 0& 0 & 1\\ 0& -1 & 0\\ -1& 0 & 0\end{bmatrix} \\ \xrightarrow{R_1\leftrightarrow R_3} \begin{bmatrix} -1& 0 & 0\\ 0& -1 & 0\\ 0& 0 & 1\end{bmatrix} \xrightarrow{-R_1, -R_2}\begin{bmatrix} 1& 0 & 0\\ 0& 1 & 0\\ 0& 0 & 1\end{bmatrix} \Rightarrow rank(A)=3 \Rightarrow \bbox[red,2pt]{\text{linearly indepentent}} \\ \det(A- \lambda I)=-(\lambda-1)^2(\lambda+1)=0 \Rightarrow \lambda=1,-1\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix} 1& 0 & 1\\ 2& -2 & 0\\ -1& 0 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+x_3=0\\ x_2+x_3=0} \\ \Rightarrow v= \begin{bmatrix} -k\\ -k \\k\end{bmatrix} ,k\in \mathbb R, 取v_1=\begin{bmatrix} -1\\ -1 \\1 \end{bmatrix}\\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix} 3& 0 & 1\\ 2& 0 & 0\\ -1& 0 & 1\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=0\\ x_3=0} \\ \Rightarrow v=\begin{bmatrix} 0\\ k \\0 \end{bmatrix} ,k\in \mathbb R, 取v_2=\begin{bmatrix} 0\\ 1 \\0 \end{bmatrix}\\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{1,-1}, \text{ eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix} -1\\ -1 \\1 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \\0 \end{bmatrix}}, \\\text{and the dimension of the eigenspace of }A=\bbox[red, 2pt] 2$$
解答:$$\\c_n= {1\over 2\pi} \int_{-\pi}^\pi e^{-x}e^{in x}dx ={1\over 2\pi} \left. \left[ {1\over in-1}e^{(in-1)x} \right] \right|_{-\pi}^\pi ={1\over 2\pi(in-1)}\left( e^{in\pi} \cdot e^{-\pi} -e^{-in\pi}\cdot e^\pi\right) \\= {1\over 2\pi(in-1)}(-1)^n(e^{-\pi}-e^\pi) \\ \Rightarrow \bbox[red, 2pt]{f(x)= \sum_{n=-\infty}^\infty c_ne^{-inx}, \text{ where }c_n={1\over 2\pi(in-1)}(-1)^n(e^{-\pi}-e^\pi)}$$
解答:$$T(x,y)=X(x)Y(y) \Rightarrow \nabla^2 T(x,y)=0 \Rightarrow X''Y+XY''=0 \\ \cases{ T(0,y)=1\\ T(\infty,y)=0\\ \frac{\partial T}{\partial y}(x,0)=0 \\ T(x,1)=0}\Rightarrow \cases{X(0)Y(y)=1\\ X(\infty)Y(y)=0\\ \Rightarrow X(x)Y'(0)=0\\ T(x)Y(1)=0} \Rightarrow \cases{X(\infty)=0\\ Y'(0)=0\\ Y(1)=0}\\ X''Y+ XY''=0 \Rightarrow {X''\over X}=-{Y''\over Y}=\mu\\ \text{Case I: }\mu=0 \Rightarrow Y''=0 \Rightarrow Y=ay+b \Rightarrow Y'=a \Rightarrow \cases{Y'(0)=0\\ Y(1)=0} \Rightarrow \cases{a=0\\ a+b=0} \\ \qquad \Rightarrow a=b=0 \Rightarrow Y=0\\ \text{Case II: }\mu
\lt 0 \Rightarrow 假設\mu= -\rho^2(\rho\gt 0) \Rightarrow Y''-\rho^2 Y=0 \Rightarrow Y= c_1e^{\rho y}+ c_2 e^{-\rho y} \\ \qquad \Rightarrow Y'= c_1\rho e^{\rho y}-c_2\rho e^{-\rho y} \Rightarrow \cases{Y'(0)=0\\ Y(1)=0} \Rightarrow \cases{(c_1-c_2)\rho =0\\ c_1e^\rho +c_2e^{-\rho}=0} \Rightarrow \cases{ c_1=c_2\\ c_1e^{2\rho}+c_2=0} \\\qquad \Rightarrow c_1(e^{2\rho}+1)=0 \Rightarrow c_1=c_2=0 \Rightarrow Y=0\\ \text{Case III: }\mu \gt 0 \Rightarrow 假設\mu=\rho^2 (\rho\gt 0) \Rightarrow Y''+\rho^2 Y=0 \Rightarrow Y=A\cos \rho x+B\sin \rho x \\ \qquad \Rightarrow Y'=-A\rho \sin \rho x+B \rho \cos \rho x \Rightarrow \cases{Y'(0)=0\\ Y(1)=0} \Rightarrow \cases{B\rho =0\\ A\cos \rho +B\sin \rho=0} \\ \qquad \Rightarrow B=0 \Rightarrow A\cos \rho=0 \Rightarrow \cos \rho=0 \Rightarrow \rho = {2n-1\over 2}\pi, n=1,2,\dots \\ \qquad \Rightarrow Y=A \cos {2n-1\over 2}\pi y, n=1,2,\dots\\ {X''\over X}=\mu= \rho^2 \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1e^{ \rho x}+c_2 e^{- \rho x}, 又X(\infty)=0 \Rightarrow c_1=0\\ \Rightarrow X=c_2e^{-\rho x}= c_2e^{-(2n-1)x/2} \Rightarrow T=XY = c_2e^{-(2n-1)x/2} A \cos {2n-1\over 2}\pi y, n=1,2,\dots \\ \Rightarrow T=\sum_{n=1}^\infty a_ne^{-(2n-1)x/2}\cos {2n-1\over 2}\pi y \\ T(0,y)=1 \Rightarrow \sum_{n=1}^\infty a_n \cos {2n-1\over 2}\pi y=1 \Rightarrow a_n= \int_0^2 \cos {2n-1\over 2}\pi y\,dy ={-4\over (2n-1)\pi} \\ \Rightarrow \bbox[red, 2pt]{T(x,y)=\sum_{n=1}^\infty {-4\over (2n-1)\pi}e^{-(2n-1)x/2}\cos {2n-1\over 2}\pi y}$$
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解題僅供參考,其他歷年試題及詳解
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