114年成大船舶機電碩士班-工程數學詳解

 國立成功大學114學年度碩士班招生考試

系所:系統及船舶機電工程學系
科目:工程數學

解答: $$\cases{P(x,y) =-2xy\\ Q(x,y) =3x^2+ y^2} \Rightarrow \cases{P_y= -2x\\ Q_x=6x} \Rightarrow P_y\ne Q_x \Rightarrow \text{Not Exact} \\ -{P_y-Q_x\over P}=-{-8x\over -2xy} =-{4\over y} \text{ depends on y only} \Rightarrow u'=-{4\over y}u \Rightarrow \text{integration factor }u=y^{-4} \\ \Rightarrow \cases{uP=-2x/y^3\\ uQ=3x^2/y^4+1/y^2} \Rightarrow \cases{(uP)_y=6x/y^4\\ (uQ)_x=6x/y^4} \Rightarrow \text{Exact}\\ \Rightarrow \Phi(x,y)=\int uP\,dx = \int uQ\, dy \Rightarrow \Phi(x,y)=\int -{2x\over y^3}\,dx = \int \left( {3x^2\over y^4}+{1\over y^2} \right)\, dy \\ \Rightarrow \Phi(x,y)=-{x^2\over y^3}+\phi(y) =-{x^2\over y^3}-{1\over y}+ \rho(x) \Rightarrow \bbox[red, 2pt]{{x^2\over y^3}+{1\over y} =c_1}$$

解答: $$A=\begin{bmatrix} 0&-1\\ -9& 0\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \lambda=3,-3\\ \cases{\lambda_1 =3 \Rightarrow v_1= \begin{bmatrix} -1\\ 3\end{bmatrix} \\ \lambda_2=-3 \Rightarrow v_2= \begin{bmatrix} 1\\ 3 \end{bmatrix}} \Rightarrow y_c=c_1\begin{bmatrix} -1\\ 3\end{bmatrix}e^{3t} +c_2 \begin{bmatrix} 1\\ 3\end{bmatrix}e^{-3t} \Rightarrow B=\begin{bmatrix} -e^{3t}& e^{-3t} \\ 3 e^{3t}& 3e^{-3t}\end{bmatrix} \\ \Rightarrow B^{-1} =\begin{bmatrix}-{1\over 2}e^{-3t} &{1\over 6}e^{-3t} \\ {1\over 2}e^{3t}& {1\over 6}e^{3t} \end{bmatrix} \Rightarrow B^{-1}g= \begin{bmatrix}-{1\over 2}e^{-3t} &{1\over 6}e^{-3t} \\ {1\over 2}e^{3t}& {1\over 6}e^{3t} \end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}-{1 \over 2}e^{-3t} \\ {1\over 2}e^{3t} \end{bmatrix} \\ \Rightarrow  \int \begin{bmatrix}-{1 \over 2}e^{-3t} \\ {1\over 2}e^{3t}\end{bmatrix}\,dt =\begin{bmatrix} {1 \over 6}e^{-3t} \\ {1\over 6} e^{3t}\end{bmatrix} \Rightarrow y_p= \begin{bmatrix} -e^{3t}& e^{-3t} \\ 3 e^{3t}& 3e^{-3t}\end{bmatrix}  \begin{bmatrix} {1 \over 6}e^{-3t} \\ {1\over 6} e^{3t} \end{bmatrix} =\begin{bmatrix} 0 \\ 1\end{bmatrix} \\ \Rightarrow y=y_c+y_p  =c_1\begin{bmatrix} -1\\ 3\end{bmatrix}e^{3t} +c_2 \begin{bmatrix} 1\\ 3 \end{bmatrix} e^{-3t}+\begin{bmatrix} 0 \\ 1\end{bmatrix} \Rightarrow \cases{y_1=-c_1e^{3t}+ c_2e^{-3t} \\y_2=3c_1e^{3t}+ 3c_2e^{-3t}+1} \\ \Rightarrow \cases{y_1(0) =-c_1+c_2=1\\ y_2(0) =3c_1+ 3c_2+1=1} \Rightarrow \cases{c_1=-1/2\\ c_2=1/2} \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1={1\over 2}e^{3t}+ {1\over 2}e^{-3t} \\y_2=-{3\over 2} e^{3t}+ {3 \over 2} e^{-3t}+1} }$$

解答: $$u(x,t) =X(x)T(t) \Rightarrow XT'=c^2X''T \Rightarrow {T'\over c^2T} ={X''\over X}=k\\ B.C.: \cases{u(0,t) =X(0)T(t) =0\\ u(L,t)= X(L)T(t) =0  }\Rightarrow\cases{X(0) =0\\ X(L)=0} \\ \textbf{Case I }k=0 \Rightarrow X=c_1x+c_2 \Rightarrow B.C.:\cases{X(0)=c_2=0\\ X(L)=c_1L+c_2=0} \Rightarrow c_1=c_2=0 \Rightarrow X=0\\ \textbf{Case II }k= \rho^2 \gt 0 \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1e^{\rho x}+ c_2e^{-\rho x}\\\qquad  \Rightarrow B.C.: \cases{X(0)= c_1+c_2=0\\ X(L)=c_1e^{\rho L} +c_2e^{-\rho L} =0} \Rightarrow c_1e^{\rho L}-c_1e^{-\rho L} =0 \Rightarrow c_1(e^{2\rho L}-1)=0 \\ \qquad \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case III }k=-\rho^2\lt 0 \Rightarrow X''+ \rho^2X =0 \Rightarrow X=c_1\cos \rho x+ c_2\sin \rho x \\\qquad \Rightarrow B.C.:\cases{ X(0)=c_1=0\\ X(L) =c_1\cos \rho L+ c_2\sin \rho L=0} \Rightarrow \sin \rho L =0 \Rightarrow \rho L=n\pi \\\qquad \Rightarrow \rho={n\pi \over L}\Rightarrow X=c_2\sin {n\pi x\over L} \\ {T'\over c^2T}=k=-\rho^2 \Rightarrow T'+\rho^2c^2 T=0 \Rightarrow T=c_3e^{-\rho^2c^2 t} =c_3e^{-n^2\pi^2 c^2 t/L^2} \\ \Rightarrow u_n(x,t) =c_2c_3 \sin{n\pi x\over L}e^{-n^2\pi^2 c^2 t/L^2} \Rightarrow u(x,t) =\sum_{n=1}^\infty a_n \sin{n\pi x\over L}e^{-n^2\pi^2 c^2 t/L^2} \\ \Rightarrow I.C.: u(x,0) =\sum_{n =1}^\infty a_n \sin{n\pi x\over L} =5\sin{3\pi x\over L} \Rightarrow \cases{a_3=5\\ a_n=0, n\ne 3} \\ \Rightarrow \bbox[red, 2pt]{u(x,t) = 5\sin{3\pi x\over L} e^{-9c^2\pi^2 t/L^2}}$$

解答:

$$\text{volume of the triangular pyramid: }{1\over 2}\cdot 2\cdot 2\cdot 4\cdot {1\over 3}={8\over 3} \\ \text{By Divergence Theorem of Gauss,} \iint_S \mathbf F\cdot \mathbf n\, dA = \iiint_R \nabla\cdot \mathbf F\, dV =\iiint_R ({\partial \over \partial x}2x+ {\partial \over \partial y}(-y))\,dV \\=\iiint_R1\,dV= \bbox[red, 2pt]{8\over 3}$$

解答: $$\text{Suppose the Fourier transform of }f(t) =\mathcal F(f(t))=\hat f(\omega)\\f(t)=e^{-3t^2} \Rightarrow f'(t)=-6te^{-3t^2} =-6t f(t) \Rightarrow \mathcal F(f'(t)) =-6 \mathcal F(tf(t)) \\ \Rightarrow i\omega \hat f(\omega) =-6i {d\over d\omega} \hat f(\omega) \Rightarrow \hat f(\omega) =c_1e^{-\omega^2/12} \Rightarrow c_1= \hat f(0) =\int_{-\infty}^\infty e^{-3t^2}\,dt \\ \Rightarrow c_1^2 =\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-3(x^2+y^2)}\,dx dy = \int_0^{2\pi} \int_0^\infty re^{-3r^2}\,drd\theta  = \int_0^{2\pi} {1\over 6}\,d\theta={\pi\over 3} \Rightarrow c_1= \sqrt{\pi\over 3} \\ \Rightarrow \hat f(\omega) = \bbox[red, 2pt]{{\sqrt{3\pi} \over 3} e^{-\omega^2/12}}$$

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