111年北科大機械碩士班乙組-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 :1120機械工程系機電整合碩士班乙組
第一節 工程數學 試題 

解答: $$\cases{P(x,y)=1+2e^{x/y} \\ Q(x,y)=2e^{x/y}(1-{x\over y})} \Rightarrow   P_y=-{2x \over y^2} e^{x/y} = Q_x  \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y)=\int (1+2e^{x/y})\,dx = \int 2e^{x/y}(1-{x\over y})\,dy\\ \Rightarrow \Phi=x+2ye^{x/y} +\rho(y) =2ye^{x/y} +\phi(x) \Rightarrow \Phi= \bbox[red,2pt]{x+2ye^{x/y} +c_1=0}$$

解答: $$\cases{P(x,y)=4y \\ Q(x,y)=x+12xy} \Rightarrow    \cases{P_y=4 \\ Q_x=1+12y} \Rightarrow P_y\ne Q_x \Rightarrow \text{ Non-exact}\\ \Rightarrow {Q_x-P_y \over P}={12y-3\over 4y}=3-{3\over 4y} \Rightarrow u'=\left(3-{3\over 4y} \right)u \Rightarrow {1\over u}du =\left(3-{3\over 4y} \right)dy \\ \Rightarrow \ln u=3y-{3\over 4}\ln y \Rightarrow u=e^{3y}y^{-3/4} \Rightarrow \cases{uP=4y^{1/4} e^{3y} \\ uQ= e^{3y}y^{-3/4}(x+12xy) } \\ \Rightarrow (uP)_y=y^{-3/4}e^{3y}+ 12y^{1/4}e^{3y} =(uQ)_x \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y)= \int 4y^{1/4} e^{3y}\,dx =\int e^{3y}y^{-3/4}(x+12xy)dy\\ \Rightarrow \Phi=4xy^{1/4} e^{3y}+\rho(y) = 4xy^{1/4} e^{3y}+\phi(x) \Rightarrow \bbox[red, 2pt]{4xy^{1/4} e^{3y}+c_1=0}$$

解答: $$y''+y=0 \Rightarrow y_h=c_1\cos x+c_2\sin x\\令\cases{y_1=\cos x\\ y_2= \sin x} \Rightarrow W=\begin{vmatrix}y_1& y_2\\ y_1'& y_2' \end{vmatrix} =1 \\\Rightarrow y_p =-\cos x\int \sin x(x\sin x)\,dx +\sin x\int \cos x(x\sin x)\,dx \\= -\cos x({1\over 4}x^2-{1\over 4}x\sin(2x)-{1\over 8}\cos (2x)) +\sin x({1\over 8}\sin(2x)- {1\over 4}x\cos(2x)) \\=-{1\over 4}x^2\cos x+{1\over 8} \cos(2x-x)+ {1\over 4}x\sin(2x-x)\\=-{1\over 4}x^2\cos x+ {1\over 8}\cos x+{1\over 4}x\sin x \Rightarrow y=y_h + y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_3\cos x+c_2\sin x-{1\over 4}x^2 \cos x +{1\over 4}x\sin x}$$

解答: $$f(t)=\begin{cases}0,& t\lt 6\\ t,& t\ge 6 \end{cases} \Rightarrow f(t)=tu(t-6)\\ \Rightarrow L\{f(t)\}= L\{tu(t-6)\} =-{d\over ds}L\{u(t-6\}  =-{d\over ds}\left({e^{-6s}\over s} \right) ={6e^{-6s} \over s} +{e^{-6s} \over s^2} \\ \Rightarrow L\{y''+4y\} =s^2Y(s)-sy(0)-y'(0)+4Y(s)=(s^2+4)Y(s) ={6e^{-6s} \over s} +{e^{-6s} \over s^2} \\ \Rightarrow Y(s) ={6e^{-6s} \over s(s^2+4)} +{e^{-6s} \over s^2(s^2+4)} \\\Rightarrow y(t)=L^{-1}\{ Y(s)\} =L^{-1} \left\{{6e^{-6s} \over s(s^2+4)} +{e^{-6s} \over s^2(s^2+4)} \right\} \\=u(t-6) \left( {3\over 2} -{3\over 2}\cos (2(t-6))\right) +u(t-6)\left( {t-6\over 4}-{1\over 8} \sin(2(t-6)) \right) \\ \Rightarrow \bbox[red, 2pt]{y(t)=u(t-6)\left( {t\over 4} -{3\over 2} \cos(2t-12)-{1\over 8} \sin(2t-16)\right)}$$

解答: $$A=\left[ \begin{matrix}-2 & 2 & -3 \\2 & 1 & -6 \\-1 & -2 & 0\end{matrix} \right] \Rightarrow \det(A-\lambda I)=-(\lambda+3)^2 (\lambda-5) =0 \Rightarrow \lambda=-3,5\\ \lambda_1=-3 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \left[ \begin{matrix}1 & 2 & -3 \\2 & 4 & -6 \\-1 & -2 & 3 \end{matrix} \right] \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_2+2x_2=3x_3 \\ \qquad \Rightarrow v=x_2\begin{bmatrix}-2\\ 1\\ 0 \end{bmatrix}+x_3 \begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}, 取v_1=\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix}, v_2= \begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix} \\ \lambda_2=5 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \left[ \begin{matrix} -7 & 2 & -3 \\2 & -4 & -6 \\ -1 & -2 & -5 \end{matrix} \right] \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+ x_3=0 \\ x_2+ 2x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}-1\\ -2\\ 1 \end{bmatrix}, 取v_3=\begin{bmatrix}-1\\ -2\\ 1 \end{bmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{-3,5}, \text{ eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix},   \begin{bmatrix} 3\\ 0\\ 1 \end{bmatrix}, \begin{bmatrix}-1\\ -2\\ 1 \end{bmatrix}}$$

解答: $$令u(x,t)= X(x)T(t)， 則 \begin{cases}\textbf{PDE} & \frac{\partial u}{\partial t} =k \frac{\partial^2 u}{\partial x^2} \Rightarrow XT'=kX''T \\\textbf{BC} & u(0,t)=u(L,t)=0 \Rightarrow X(0)=X(L)=0 \\ \textbf{IC}& u(x,0)=A \end{cases}\\ XT'=kX''T \Rightarrow {T'\over kT}= {X''\over X}= \lambda\\ \textbf{Case I }\lambda=0.  \Rightarrow X''=0 \Rightarrow X=c_1x+ c_2 \Rightarrow \textbf{BC}:\cases{X(0)=0\\ X(L)=0} \Rightarrow \cases{c_2=0 \\ c_1L+c_2=0} \\\qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0 \\ \textbf{Cases II } \lambda\gt 0. 假設\lambda =\rho^2 (\rho \gt 0) \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x} +c_2 e^{-\rho x} \\ \qquad \textbf{BC}:\cases{X(0)=0\\ X(L)=0} \Rightarrow \cases{c_1+c_2=0 \\ c_1e^{\rho L} + c_2 e^{-\rho L}=0} \Rightarrow c_1e^{\rho L} - c_1 e^{-\rho L}=0 \\ \qquad  \Rightarrow c_1(e^{2\rho L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0 \\ \textbf{Cases III } \lambda\lt 0. 假設\lambda =-\rho^2 (\rho \gt 0) \Rightarrow X''+\rho^2 X=0 \Rightarrow X=c_1 \cos(\rho x) +c_2 \sin(\rho x)  \\ \qquad \textbf{BC}:\cases{X(0)=0\\ X(L)=0} \Rightarrow \cases{c_1 =0 \\ c_1\cos(\rho L) + c_2 \sin(\rho L)=0} \Rightarrow c_2\sin(\rho L)=0\\ \qquad \Rightarrow \sin(\rho L)=0 \Rightarrow \rho = {n\pi \over L} \Rightarrow X= c_2\sin(n\pi x/L), n\in \mathbb N \\ \Rightarrow T'+\rho^2kT=0 \Rightarrow T=c_3e^{-\rho^2 kt} =c_3e^{-n^2\pi^2kt/L^2} \\\Rightarrow u(x,t)=XT= c_2c_3e^{-n^2\pi^2kt/L^2} \sin(n\pi x/L) =\sum_{n=1}^\infty a_n e^{-n^2\pi^2kt/L^2} \sin(n\pi x/L) \\ \textbf{IC: }u(x,0)=\sum_{n=1}^\infty a_n   \sin(n\pi x/L) =A \Rightarrow a_n={2\over L} \int_0^L A\sin(n\pi x/L)dx \\=-{2A\over n\pi} \left[ \cos(n\pi)-1\right]= {2A\over n\pi}(1-(-1)^n) \\ \Rightarrow \bbox[red, 2pt]{u(x,t)=\sum_{n=1}^\infty {2A\over n\pi}(1-(-1)^n) e^{-n^2\pi^2kt/L^2} \sin(n\pi x/L)}$$

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

解題僅供參考, 其他歷年試題及詳解