115年台科大機械碩士班丙組-工程數學詳解

國立立臺灣科技大學115學年度碩士班招生

系所組別:機械系碩士班丙組
科目:工程數學

解答:$$\textbf{(a) }y'-y=e^{2x} \Rightarrow y'e^{-x}-ye^{-x}=e^x \Rightarrow \left( ye^{-x} \right)' =e^x \Rightarrow ye^{-x} =\int e^x\,dx =e^x+c_1 \\ \qquad \Rightarrow \bbox[red, 2pt]{y=e^{2x}+c_1e^x} \\\textbf{(b) }y''-2y'-2y=0 \Rightarrow \lambda^2-2\lambda-2=0 \Rightarrow \lambda =1\pm \sqrt 3 \Rightarrow \bbox[red, 2pt]{y=c_1 e^{(1+\sqrt 3)x} +c_2e^{(1-\sqrt 3)x}}$$

解答:$$\textbf{(a) }L\{f(t)\} = \bbox[red, 2pt]{\int_0^\infty f(t)e^{-st} \,dt}\\ \textbf{(b) }L\{t^2f(t)\} ={d^2 \over ds^2}F(s) \Rightarrow L\{t^2 \sin(\omega t)\} ={d^2\over ds^2} L\{\sin(\omega t)\} ={d^2 \over ds^2} \left( {\omega\over s^2+\omega^2} \right) \\\qquad = {d\over ds} \left( {-2\omega s\over (s^2+\omega^2)^2} \right) ={6\omega s^2-2\omega^3\over (s^2+ \omega^2)^3} = \bbox[red, 2pt]{2\omega(3s^2-\omega^2) \over (s^2+\omega^2)^3}\\ \textbf{(c) }L^{-1} \left\{{s \over (s+2)^2 (s^2+4s+8)} \right\} =L^{-1} \left\{ {1\over 4(s+2)}-{1\over 2(s+2)^2}-{s\over 4(s^2+4s+8)} \right\}\\= L^{-1} \left\{ {1\over 4(s+2)}-{1\over 2(s+2)^2}- \left( {1\over 4}\cdot {s+2\over (s+2)^2+4}-{1\over 2}\cdot {1\over (s+2)^2+4} \right) \right\}\\={1\over 4}e^{-2t}-{1\over 2}te^{-2t}-{1\over 4}e^{-2t}(\cos(2t)-\sin(2t)) \\ \Rightarrow L^{-1} \left\{{se^{-2s} \over (s+2)^2 (s^2+4s+8)} \right\} =u(t-2)L^{-1} \left\{{s \over (s+2)^2 (s^2+4s+8)} \right\}(t-2) \\  \\=u(t-2) \left( {1\over 4}e^{-2(t-2)} -{1\over 2}(t-2)e^{-2(t-2)} -{1\over 4}e^{-2(t-2)} (\cos(2(t-2))-\sin(2(t-2)))\right) \\= \bbox[red, 2pt]{u(t-2)e^{-2(t-2)} \left[{1\over 4}-{1\over 2}(t-2)-{1\over 4}(\cos(2(t-2))-\sin(2(t-2)))  \right]}$$

解答:$$C= C_1\cup C_2 \cup C_3 \text{ where }\cases{C_1= \{(t,t,0) \mid 0\le t\le 1\} \\ C_2=\{(1,1,t) \mid 0\le t\le 1\} \\C_3= \{(1-t,1-t,1-t) \mid 0\le t\le 1\}} \Rightarrow \oint_C \vec F\cdot d\vec r \\=\int_C (xdx-zdy+2ydz) = \int_0^1 t\,dt +\int_0^1 2\,dt + \int_0^1 (2t-2)\,dt={1\over 2}+2 -1=\bbox[red, 2pt]{3\over 2}$$

解答:$$\textbf{(a) }u(x,t)=X(x)T(t) \Rightarrow XT'-kX''T=0 \Rightarrow {T'\over kT} ={X''\over X}=-\lambda \Rightarrow \cases{X''+\lambda X=0\\ T'+\lambda kT=0} \\\text{Solving for }X(x): B.C. \cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \Rightarrow \lambda=\beta^2 \gt 0 \\  \qquad \Rightarrow X= C_1\cos(\beta x)+C_2 \sin(\beta x) \Rightarrow X(0)=C_1=0 \Rightarrow X(L)=C_2\sin (\beta L)=0 \\\qquad \Rightarrow \beta_n={n\pi\over L} \Rightarrow \lambda_n= \left( {n\pi\over L} \right)^2 \Rightarrow X_n=\sin {n\pi x\over L}, n=1,2,\dots \\ \text{Solving for }T(t): T'+\lambda_n kT=0 \Rightarrow T_n =e^{-k({n\pi\over L})^2t} ,n=1,2,\dots \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= \sum_{n=1}^\infty B_n \sin{n\pi x\over L}e^{-k({n\pi\over L})^2t}} \\\textbf{(b) } u(x,0)=A= \sum_{n=1}^\infty B_n \sin{n\pi x\over L} \Rightarrow B_n={2\over L} \int_0^L A\sin{n\pi x\over L}\,dx = \bbox[red, 2pt]{{2A\over n\pi}(1-(-1)^n)}$$

解答:$$\cases{x_1'(t)= 3x_1(t)-x_2(t)-x_3(t) \\ x_2'(t)= x_1(t)+x_2(t)-x_3(t)+t \\ x_3'(t)= x_1(t)-x_2(t)+ x_3(t)+ 2e^t}  \Rightarrow \mathbf x'=A\mathbf x+ b, \text{ where }A= \begin{bmatrix}3 &-1&-1\\1&1& -1\\1&-1&1 \end{bmatrix},b= \begin{bmatrix}0\\t\\2e^t \end{bmatrix}\\ \Rightarrow \det(A-\lambda I) =-(\lambda-1)(\lambda-2)^2 \Rightarrow \cases{\lambda=1 \Rightarrow \text{eigenvector }v_1= \begin{bmatrix}1 \\1\\1 \end{bmatrix} \\ \lambda=2 \Rightarrow \text{eigenvector }v_2 =   \begin{bmatrix}1\\1\\0 \end{bmatrix}\text{ and }v_3= \begin{bmatrix}1 \\0\\1 \end{bmatrix}} \\ \Rightarrow x_h(t) =c_1 e^t \begin{bmatrix}1\\1\\1 \end{bmatrix} +c_2 e^{2t} \begin{bmatrix}1\\1\\0 \end{bmatrix} + c_3 e^{2t} \begin{bmatrix}1\\ 0\\1 \end{bmatrix} \Rightarrow B = \begin{bmatrix}e^t& e^{2t} & e^{2t} \\ e^t& e^{2t} & 0\\e^t& 0& e^{2t}\end{bmatrix} \Rightarrow B^{-1} = \begin{bmatrix}-e^{-t }& e^{-t} & e^{-t} \\ e^{-2t}& 0& -e^{-2t} \\ e^{-2t} & -e^{-2t} & 0\end{bmatrix} \\ \Rightarrow B^{-1} b= \begin{bmatrix}te^{-t}+2 \\-2e^{-t}\\-te^{-2t} \end{bmatrix} \Rightarrow \int \begin{bmatrix}te^{-t}+2 \\-2e^{-t}\\-te^{-2t}  \end{bmatrix}\,dt = \begin{bmatrix}2t-e^{-t}(t+1) \\2e^{-t} \\ {1\over 4}e^{-2t}(2t+1)\end{bmatrix}\\ \Rightarrow x_p(t)= \begin{bmatrix}e^t& e^{2t} & e^{2t} \\ e^t& e^{2t} & 0\\e^t& 0& e^{2t}\end{bmatrix} \begin{bmatrix}2t-e^{-t}(t+1) \\2e^{-t} \\ {1\over 4}e^{-2t}(2t+1)\end{bmatrix} = \begin{bmatrix}-{1\over 2}t+2te^t+ 2e^t-{3\over 4} \\ -t+ 2te^t+2e^t-1\\ -{1\over 2}t+2te^t-{3\over 4}\end{bmatrix} \Rightarrow x=x_h+ x_p \\ \Rightarrow x(0)= \begin{bmatrix}1\\2 \\-2 \end{bmatrix} = \begin{bmatrix}c_1+c_2+c_3+2-{3\over 4} \\ c_1+c_2+2-1\\ c_1+c_3-{3\over 4} \end{bmatrix}  \Rightarrow \cases{c_1 =0\\ c_2=1\\ c_3=-5/4} \\\Rightarrow  \bbox[red, 2pt]{\cases{x_1(t)=2(t+1)e^t-{1\over 4}e^{2t}-{1\over 2}t-{3\over 4} \\x_2(t) =2(t+1)e^t+e^{2t}-t-1\\ x_3(t) =2te^t-{5\over 4}e^{2t} -{1\over 2}t-{3\over 4}}}$$

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解題僅供參考，碩士班歷年試題及詳解