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:機械工程碩士班甲組、乙組、丙組、丁組
科目名稱:工程數學 

 

解答:$$\textbf{(a) }xy''-y'=2x^2 \Rightarrow {1\over x}y''-{1\over x^2}=2 \Rightarrow \left( {1\over x}y'\right)'=2 \Rightarrow {1\over x}y'=2x+c_1 \\ \quad \Rightarrow y'=2x^2+c_1x \Rightarrow y={2\over 3}x^3+{1\over 2}c_1x^2+c_2 \Rightarrow \cases{y(0)=c_2=0\\ y(1)={2\over 3}+{1\over 2}c_1+c_2=1} \\ \quad \Rightarrow \cases{ c_1={2\over 3}\\c_2=0} \Rightarrow \bbox[red, 2pt]{y= {2\over 3}x^3+ {1\over 3}x^2} \\\textbf{(b) } y''+5y'+6y=0 \Rightarrow \lambda^2+ 5\lambda+6=0 \Rightarrow \lambda=-2,-3 \Rightarrow y_h=c_1e^{-2x} +c_2e^{-3x} \\ \quad y_p=A xe^{-2x} \Rightarrow y_p'=Ae^{-2x} -2Axe^{-2x} \Rightarrow y_p''=-4Ae^{-2x}  + 4Axe^{-2x}\\\quad \Rightarrow y_p''+5y_p'+6y_p= Ae^{-2x} =e^{-2x} \Rightarrow A=1\Rightarrow   y_p= xe^{-2x} \\\quad \Rightarrow y=y_h+ y_p =c_1e^{-2x} +c_2e^{-3x} + xe^{-2x}  \Rightarrow \cases{y(0) =c_1+c_2 =0\\ y(1) =c_1e^{-2} +c_2e^{-3}+ e^{-2}=e^{-3}}  \\\quad \Rightarrow \cases{c_1=-1\\ c_2=1} \Rightarrow \bbox[red, 2pt]{y=-e^{-2x} +e^{-3x}+ xe^{-2x}}$$

解答:$$f(x)=\int_0^x e^{-\tau} \cos \tau\, d\tau = \left. \left[ {1\over 2}e^{-\tau} (\sin \tau- 
\cos \tau) \right] \right|_0^x ={1\over 2}e^{-x}(\sin x-\cos x)+{1\over 2} \\ \Rightarrow L\{f(x)\} ={1\over 2}L\{ e^{-x}(\sin x-\cos x)\} +{1\over 2}L\{1\} ={1\over 2}\left({1\over (s+1)^2+1}- {s+1\over (s+1)^2+1}\right)+{1\over 2s} \\ \Rightarrow L\{f(x)\} =\bbox[red, 2pt]{{1\over 2} \left( {1\over s}-{s\over (s+1)^2+1}\right)}$$

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

解答:$$A=\begin{bmatrix} 0&-2\\ 1& 3\end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-3\lambda+2 =0 \Rightarrow \lambda=1,2 \\ \Rightarrow p(\lambda) =e^\lambda=a\lambda +b \Rightarrow \cases{p(1) =e=a+b\\ p(2)=e^2=2a+b} \Rightarrow \cases{a=e^2-e\\ b=-e^2+2e} \Rightarrow e^\lambda=(e^2-e)\lambda+(-e^2+2e) \\ \Rightarrow e^A= (e^2-e)A+(-e^2+2e) I=\begin{bmatrix} 0&-2(e^2-e)\\ (e^2-e) & 3 (e^2-e)\end{bmatrix} +\begin{bmatrix} -e^2+2e& 0\\0& -e^2+2e\end{bmatrix} \\ \Rightarrow e^A=\bbox[red, 2pt]{\begin{bmatrix} -e^2+2e& -2e^2+2e\\ e^2-e& 2e^2-e\end{bmatrix}}$$

解答:$$u(x,t)= v(x,t)=ax+b, \text{where }a\text{ and }b\text{ are constant} \Rightarrow \cases{u(-1,t) =v(-1,t)-a+b=2\\ u(1,t)=v(1,t)+a+b=4} \\ \Rightarrow \cases{v(-1,t)=0\\ v(1,t) =0\\ a=1\\ b=3} \Rightarrow u(x,t)=v(x,t)+x+3 \Rightarrow \cases{\frac{\partial v(x,t)}{\partial t} = \frac{\partial^2 v(x,t)}{\partial x^2}\\v(-1,t)=0\\ v(1,t) =0\\ v(x,0)= \sin(2\pi x)} \\\text{Suppose }v(x,t)=X(x+1)T(t), \text{ then we have } \cases{v(-1,t)= X(0)T(t)=0\\ v(1,t)=X(2)T(t)=0} \ \Rightarrow \cases{X(0)=0\\ X(2)=0} \\ \qquad \text{ and }{X''(x+1) \over X(x+1)}={T'(t) \over T(t)} =\lambda \\\textbf{Case I }\lambda=0 \Rightarrow X''(x+1)=0 \Rightarrow X(x+1) =c_1x+c_2 \Rightarrow X(x)=c_1(x-1)+c_2 =c_1x+c_3\\ \qquad \Rightarrow \cases{X(0)=c_3=0\\ X(2) =2c_1+c_3=0} \Rightarrow \cases{c_1=0\\ c_3=0} \Rightarrow X=0 \Rightarrow X(x+1)=0 \Rightarrow v=0 \\ \textbf{Case II }\lambda \gt 0 \Rightarrow \lambda=\rho^2 \Rightarrow X^2(x+1)- \rho^2 X(x+1)=0 \Rightarrow X(x+1)= c_1e^{\rho x} +c_2 e^{-\rho x} \\ \qquad \Rightarrow X(x)=c_1e^{\rho(x-1)} +c_2e^{-\rho(x-1)} =c_3e^{\rho x}+ c_4 e^{-\rho x} \Rightarrow \cases{X(0)=c_3+c_4=0\\ X(2)= c_3e^{2\rho} +c_4e^{-2\rho } =0} \\\qquad \Rightarrow c_3e^{2\rho}-c_3e^{-2\rho}=0 \Rightarrow c_3(e^{4\rho}-1)=0 \Rightarrow c_3=0 \Rightarrow c_4=0 \Rightarrow X=0 \Rightarrow v=0 \\\\ \textbf{Case III }\lambda \lt 0 \Rightarrow \lambda=-\rho^2 \Rightarrow X^2(x+1)+ \rho^2 X(x+1)=0 \Rightarrow X(x+1)= c_1\cos(\rho x) +c_2 \sin(\rho x) \\ \qquad \Rightarrow X(x)=c_1 \cos(\rho(x-1))+ c_2\sin(\rho(x-1)) = c_3\cos(\rho x)+ c_4\sin(\rho x) \\\qquad \Rightarrow \cases{X(0)=c_3=0\\ X(2)=c_3\cos(2\rho) +c_4\sin(2\rho)=0} \Rightarrow \sin(2\rho)=0 \Rightarrow 2\rho =n\pi \Rightarrow \rho ={n\pi \over 2}\\ \qquad \Rightarrow X_n(x) = B_n \sin{n\pi x\over 2},n\in \mathbb N \Rightarrow T'+ \rho^2 T=0 \Rightarrow T=c_5e^{-\rho^2 t} \Rightarrow T_n(t)= A_ne^{-n^2\pi^2 t/4} \\ \Rightarrow v(x,t)=X(x+1)T(t) =\sum_{n=1}^\infty C_n \sin{n\pi(x+1)\over 2} e^{-n^2\pi^2 t/4} \Rightarrow v(x,0)= \sum_{n=1}^\infty C_n \sin{n\pi(x+1)\over 2} =\sin(2\pi x) \\ \Rightarrow \cases{C_4=1\\ C_n=0,n\ne 4} \Rightarrow v(x,t)= \sin (2\pi(x+1)) e^{-4\pi^2 t} =\sin(2\pi x)e^{-4\pi^2 t} \\ \Rightarrow u(x,t)=v(x,t)+x+3 \Rightarrow \bbox[red, 2pt]{u(x,t)= \sin(2\pi x)e^{-4\pi^2 t} +x+3}$$

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解題僅供參考，碩士班歷年試題及詳解