111年北科大機械碩士班甲組-工程數學詳解

國立臺北科技大學111學年度碩士班招生考試

系所組別 :11ll、 l112機械工程系機電整合碩士班甲組
第一節 工程數學 試題 

解答：$$\textbf{(1)}\;2\sin ydx+ \cos y dy =0 \Rightarrow {\cos y\over \sin y}dy =-2dx \Rightarrow \ln(\sin y)=-2x+c_1 \Rightarrow \sin y=c_2e^{-2x}\\ y(0)={\pi \over 2} \Rightarrow \sin({\pi \over 2})=1=c_2 \Rightarrow \sin y= e^{-2x} \Rightarrow \bbox[red, 2pt]{y=\sin^{-1} e^{-2x}} \\\textbf{(2)}\;xy'+y-e^x=0 \Rightarrow (xy)'=e^x \Rightarrow xy=e^x +c_1\\ \quad y(1)=e代入上式 \Rightarrow e=e+c_1 \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt]{y={1\over x}e^x}$$

解答：$$L\{y''\} +4L\{y'\} +3L\{y \} =L\{e^t\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0)) +3Y(s)= {1\over s-1} \\ \Rightarrow (s^2+4s +3)Y(s)-2={1\over s-1} \Rightarrow Y(s)={1\over (s+3)(s+1)(s-1)}+ {2\over (s+3)(s+1)} \\ =\left({1\over 8(s+3)}-{1\over 4(s+1)} +{1\over 8(s-1)} \right)+\left( -{1\over (s+3) } +{1\over (s+1)}\right) \\=-{7\over 8} \cdot {1\over s+3} +{3\over 4}\cdot {1\over s+1}+{1\over 8}\cdot {1\over s-1} \Rightarrow y(t)=L^{-1}\{Y(s)\} \\ \Rightarrow \bbox[red, 2pt]{y(t)=-{7 \over 8}e^{-3t}+{3 \over 4}e^{-t}+{1\over 8}e^t}$$

解答：$$\textbf{(1)}\; 令\cases{P(x,y)=3x^2(y^2-4y) \\ Q(x,y)= 2x^3y-4x^3} \Rightarrow \frac{\partial }{\partial y}P =3x^2(2y-4) =\frac{\partial }{\partial x}Q \Rightarrow \vec F \text{ is a conservative field}, \\\qquad \bbox[red, 2pt]{Q.E.D} \\ \textbf{(2)}\; \phi=\int P\,dx =\int Q\,dy \Rightarrow \phi =\int 3x^2(y^2-4y)\,dx =\int 2x^3y-4x^3\,dy\\ \quad =x^3y^2-4x^3y+ \rho(y)= x^3y^2-4x^3y+ \delta(x) \Rightarrow \bbox[red, 2pt]{\phi=x^3y^2-4x^3y+c_1}\\ \textbf{(3)}\;  \int_C \vec F(r)\cdot d\vec r = \phi(2,2)-\phi(-1,1) =\bbox[red, 2pt]{-35}$$

解答：$$\textbf{(1)}\; \cases{2x_1+ 6x_2+x_3=7 \\ x_1+2x_2 -x_3=-1 \\ 5x_1+7x_2-4x_3=9} \Rightarrow \left[ \begin{array}{rrr|r }2& 6 & 1& 7\\ 1& 2& -1 & -1\\ 5& 7& -4& 9\end{array} \right] \xrightarrow{-2R_2+R_1\to R_1,-5R_2+R_3\to R_3} \\\left[ \begin{array}{rrr|r }0& 2 & 3& 9\\ 1& 2& -1 & -1\\ 0& -3& 1& 14\end{array} \right] \xrightarrow{R_1/2\to R_1}\left[ \begin{array}{rrr|r }0& 1 & 3/2& 9/2\\ 1& 2& -1 & -1\\ 0& -3& 1& 14\end{array} \right] \xrightarrow{-2R_1+R_2\to R_2,3R_1+R_3 \to R_3} \\\left[ \begin{array}{rrr|r }0& 1 & 3/2& 9/2\\ 1& 0& -4 & -10\\ 0& 0& 11/2& 55/2\end{array} \right] \xrightarrow {R_1\leftrightarrow R_2,2R_3/11\to R_3} \left[ \begin{array}{rrr|r } 1& 0& -4 & -10\\0& 1 & 3/2& 9/2\\  0& 0& 1& 5\end{array} \right] \\ \Rightarrow \cases{x_1-4x_3=-10\\ x_2+3x_3/2=9/2\\ x_3=5} \Rightarrow \bbox[red, 2pt]{\cases{x_1=10\\ x_2= -3\\ x_3=5}} \\\textbf{(2)}\; A=\begin{bmatrix}2& 6 & 1\\ 1& 2& -1\\ 5& 7& -4 \end{bmatrix} \Rightarrow A^{-1} =\left[\begin{matrix} \frac{1}{11} & \frac{-31}{11} & \frac{8}{11} \\ \frac{1}{11} & \frac{13}{11} & \frac{-3}{11} \\ \frac{3}{11} & \frac{-16}{11} & \frac{2}{11}\end{matrix} \right] \\\qquad \Rightarrow \textbf{x} =A^{-1}\textbf{c} = \left[\begin{matrix} \frac{1}{11} & \frac{-31}{11} & \frac{8}{11} \\ \frac{1}{11} & \frac{13}{11} & \frac{-3}{11} \\ \frac{3}{11} & \frac{-16}{11} & \frac{2}{11}\end{matrix} \right] \begin{bmatrix}7\\ -1 \\ 9 \end{bmatrix} = \begin{bmatrix}10\\ -3 \\ 5 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{x_1=10\\ x_2=-3\\ x_3=5}}$$

解答：$$u(x,t)= X(x)T(t) \Rightarrow \begin{cases}\textbf{PDE}& X(x)T''(t)=X''(x)T(t) \\\textbf{BC}& X(0)T(t)=X(\pi)T(t)=0 \Rightarrow X(0)=X(\pi)=0\\ \textbf{IC}& X(x)T'(0)=0 \Rightarrow T'(0)=0 \end{cases} \\ XT''=X''T \Rightarrow {T''\over T}={X''\over X}= \lambda \\ \textbf{Case I }\lambda=0. \;X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \textbf{BC}:\cases{X(0)=0\\ X(\pi)=0} \Rightarrow \cases{c_2=0 \\ c_1\pi +c_2=0} \\ \qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0\\\textbf{Case II }\lambda\gt 0. \; 令\lambda =\rho^2 (\rho \gt 0) \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x} +c_2e^{-\rho x}\\ \qquad \Rightarrow \textbf{BC}:\cases{X(0)=0\\ X(\pi)=0} \Rightarrow \cases{c_1+c_2=0 \\ c_1e^{\rho \pi} +c_2e^{-\rho \pi}=0} \Rightarrow c_1e^{\rho \pi}-c_1e^{-\rho \pi}=0\\ \qquad \Rightarrow c_1(e^{2\rho \pi}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case III }\lambda \lt 0. 令\lambda=-\rho^2(\rho \gt 0) \Rightarrow X''+\rho^2 X=0 \Rightarrow X= c_1 \cos(\rho x)+ c_2\sin(\rho x)\\ \qquad \textbf{BC}: \cases{X(0)=0\\ X(\pi)=0} \Rightarrow \cases{c_1=0\\ c_2\sin( \rho \pi)=0} \Rightarrow \sin(\rho \pi)=0 \Rightarrow \rho=n \Rightarrow X=c_2 \sin(n x), n\in \mathbb N\\ \Rightarrow T''+\rho^2 T=0 \Rightarrow T= c_3\cos(\rho t)+ c_4\sin(\rho t) \Rightarrow T'=\rho c_4\cos(\rho t)-\rho c_3 \sin(\rho t)\\ \textbf{IC}: T'(0)=0 \Rightarrow \rho c_4=0 \Rightarrow c_4=0 \Rightarrow T=c_3 \cos(n t),n\in \mathbb N \cup \{0\}\\ \Rightarrow u(x,t)=X(x)T(t)= \sum_{n=0}^\infty a_n \cos(nt)\sin(nx) \Rightarrow u(x,0)= \sum_{n=0}^\infty a_n  \sin(nx) =\sin x-\sin(2x) \\ \Rightarrow \cases{a_0=0\\ a_1=1\\ a_2=-1 \\a_k=0,k\ge 3} \Rightarrow \bbox[red, 2pt] {u(x,y)=\cos(t)\sin(x)-\cos(2t)\sin(2x)}$$

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