110年台科大機械碩士班-工程數學詳解

 國立臺灣科技大學110學年度碩士班招生考試

系所組別:機械工程系碩士班
科目:工程數學

解答: $$\textbf{(a) }xy''-y'=2x^2 \Rightarrow {1\over x}y''-{1\over x^2}y'=2 \Rightarrow \left( {1\over x}y'\right)'=2 \Rightarrow {1\over x}y'=2x+c_1 \Rightarrow y'=2x^2+c_1 x \\ \quad \Rightarrow y={2\over 3}x^3+c_2x^2+c_3 \Rightarrow \cases{y(0)=c_3=0\\ y(1)=2/3+c_2+c_3=1} \Rightarrow c_2={1\over 3} \\\quad \Rightarrow \bbox[red, 2pt]{y={2\over 3}x^3+{1\over 3}x^2} \\\textbf{(b) }y''+5y'+6y=0 \Rightarrow \lambda^2+5\lambda+6=0 \Rightarrow (\lambda+3)(\lambda+2)=0 \Rightarrow \lambda=-2,-3 \Rightarrow y_h=c_1e^{-2x} +c_2e^{-3x} \\\quad \text{Using variations of parameters, let }\cases{y_1=e^{-2x} \\y_2= e^{-3x}} \Rightarrow W(x)= \begin{vmatrix}y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix}e^{-2x} & e^{-3x}\\ -2e^{-2x} & -3e^{-3x} \end{vmatrix} =-e^{-5x} \\ \quad \Rightarrow y_p = -e^{-2x}\int{ e^{-3x} \cdot e^{-2x}\over -e^{-5x}}\,dx +e^{-3x }\int{ e^{-2x} \cdot e^{-2x}\over -e^{-5x}} \,dx =e^{-2x} \int 1\,dx -e^{-3x}\int e^x\,dx \\\quad \Rightarrow y_p=xe^{-2x}-e^{-2x} \Rightarrow y=y_h+y_p \Rightarrow y=c_3e^{-2x} +c_2e^{-3x} +xe^{-2x} \\ \quad \Rightarrow \cases{y(0)=c_3+c_2=0\\ y(1)= (c_3+1)e^{-2} +c_2e^{-3} =e^{-3}} \Rightarrow \cases{c_2=1\\ c_3=-1} \Rightarrow \bbox[red, 2pt]{y=-e^{-2x}+e^{-3x}+xe^{-2x}}$$

解答: $$L\{f(t)\} ={L\{e^{-t} \cos t\} \over s} ={1\over s} \cdot {s+1\over (s+1)^2+1} = \bbox[red, 2pt]{s+1\over s(s^2+2s+2)}$$

解答: $$f(t)+ 2\int_0^t f(\tau) \cos(t-\tau)\,d\tau =4e^{-t}+ \sin t\\ \Rightarrow L\{f(t)\}+ 2L \left\{\int_0^t f(\tau) \cos(t-\tau)\,d\tau \right\} =4L\{e^{-t}\}+ L\{\sin t\} \\ \Rightarrow F(s)+2 L\{f(t)\} L\{\cos t\} =4\cdot {1\over s+1}+ {1\over s^2+1} \\ \Rightarrow F(s)+2F(s)\cdot {s\over s^2+1} = {4\over s+1}+ {1\over s^2+1}\\ \Rightarrow F(s)=\left( {4\over s+1}+ {1\over s^2+1}\right) \cdot {s^2+1\over (s+1)^2} ={4(s^2+1) \over (s+1)^3} +{1\over (s+1)^2} ={4s^2+s+5 \over (s+1)^3} \\ \Rightarrow f(t)=L^{-1} \{F(s) \}=L^{-1}\left\{ {4s^2+s+5 \over (s+1)^3} \right\} =L^{-1} \left\{ {4\over s+1}-{7\over (s+1)^2} +{8\over (s+1)^3}\right\}\\ \Rightarrow \bbox[red, 2pt]{f(t)= 4e^{-t}-7te^{-t} +4t^2 e^{-t}}$$

解答: $$A=\begin{bmatrix}0&-2\\ 1& 3 \end{bmatrix}  \Rightarrow \det(A-\lambda I)= (\lambda-1)(\lambda -2)=0 \Rightarrow \lambda=1,2\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} -1 & -2 \\1 & 2\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1+2x_2=0 \Rightarrow v= x_2 \begin{pmatrix} -2\\ 1\end{pmatrix} \\\qquad \text{choose }v_1= \begin{pmatrix} -2\\ 1\end{pmatrix}\\ \lambda_2=2  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -2 & -2 \\1 & 1\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=-x_2 \Rightarrow v= x_2 \begin{pmatrix} -1\\ 1\end{pmatrix} \\\qquad \text{choose }v_2= \begin{pmatrix} -1\\ 1\end{pmatrix} \\ \Rightarrow A=\begin{bmatrix} -2& -1\\ 1& 1\end{bmatrix} \begin{bmatrix} 1& 0\\ 0& 2 \end{bmatrix} \begin{bmatrix} -2& -1\\ 1& 1\end{bmatrix}^{-1} \Rightarrow e^A= \begin{bmatrix} -2& -1\\ 1& 1\end{bmatrix} \begin{bmatrix} e& 0\\ 0& e^2 \end{bmatrix} \begin{bmatrix} -2& -1\\ 1& 1\end{bmatrix}^{-1} \\= \bbox[red, 2pt]{\begin{bmatrix} 2e-e^2& 2e-2e^2\\ -e+e^2& -e+2e^2\end{bmatrix}}$$

解答: $$u(x,t)=v(x,t)+u_{ss}(x)\\ u_{ss}x= \alpha x+\beta \Rightarrow \cases{u(-1,t)=-\alpha+\beta=2\\ u(1,t)=\alpha+\beta=4} \Rightarrow \cases{\alpha=1\\ \beta=3} \Rightarrow u_{ss}(x)= x+3\\ \Rightarrow u(x,0)=3+x+\sin(2\pi x)-u_{ss}(x)= \sin(2\pi x) \\ \Rightarrow v(x,t)= \sum_{n=1}^\infty b_n e^{-n^2\pi^2 t/4} \sin {n\pi\over 2}(x+1) \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= x+3+ \sum_{n=1 }^\infty b_n e^{-n^2\pi^2 t/4} \sin {n\pi\over 2}(x+1) ,  \text{where }b_n= \int_{-1}^1 \sin(2\pi x)\sin{n\pi\over 2}(x+1)\,dx }$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解