114年成大工程科學碩士班-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:工程科學系
科目:工程數學

解答: $$y''-3x^{-1}y'+4x^{-2}y=2 \Rightarrow x^2y''-3xy'+4y=2x^2\\y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\\Rightarrow x^2y''-3x y'+4 y=m(m-1)x^{m}-3mx^{m}+4x^{m} =(m^2-4m+4)x^{m}=0 \\ \Rightarrow (m-2)^2=0 \Rightarrow m=2 \Rightarrow y_h= c_1x^2+ c_2x^2\ln x\\ y_p=A(\ln x)^2 x^2 \Rightarrow y_p'= 2A(\ln x)^2x+ 2A(\ln x)x \Rightarrow y_p''=2A(\ln x)^2 +6A(\ln x)+  2A \\ \Rightarrow x^2y_p''-3xy_p'+4y_p =2Ax^2 = 2x^2 \Rightarrow A=1 \Rightarrow y_p=(\ln x)^2 x^2 \\ \Rightarrow y=y_h+ y_p \Rightarrow y= c_1x^2+ c_2x^2\ln x+x^2(\ln x)^2,故選\bbox[red, 2pt]{(D)}$$

解答: $$L\{ \cos(kt) \delta(t-k\pi)\} = \int_0^\infty e^{-st} \cos(kt) \delta(t-k\pi)\,dt =e^{-k\pi s }\cos(k^2 \pi) = -e^{-3\pi s} \text{ for }3\pi \le t\lt 4\pi  \\ L\{y'(t)\} +9L\{ \int_0^t y(\tau )\,d \tau\} =sY(s)+0 +9{Y(S)\over s}=-e^{-3\pi s} \Rightarrow Y(s)=-{se^{-3\pi s}\over s^2+9} \\ \Rightarrow y(t)=L^{-1}\left\{-{se^{-3\pi s}\over s^2+9} \right\} = -u(t-3\pi )\cos(3(t-3\pi)) =\cos 3t, \text{ for }3\pi\le t\lt 4\pi, 故選\bbox[red, 2pt]{(F)}$$

解答: $$S_N(x)={1\over 2\pi}+ {1\over \pi} \sum_{n=1}^N  \cos(nx) ={1\over 2\pi}+ {1\over \pi}\left( {\sin(N+{1\over 2}x) \over 2\sin{1\over 2}x} -{1\over 2}\right) ={1\over 2\pi} \cdot {\sin(N+{1\over 2}x) \over 2\sin{1\over 2}x},故選\bbox[red, 2pt]{(B)}$$

解答: $$u(x,t) =X(x)T(t) \Rightarrow XT''=c^2 X''T \Rightarrow {X''\over X} ={T''\over c^2 T} =\lambda= -\alpha^2 \lt 0 \\ \Rightarrow X''+\alpha^2X=0 \Rightarrow X=c_1\cos \alpha x+ c_2\sin \alpha x \\ \cases{u(0,t)= X(0)T(t)=0\\ u(L,t)= X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \Rightarrow \cases{c_1=0\\ c_1\cos \alpha L+c_2\sin \alpha L=0} \Rightarrow \sin \alpha L=0\\ \Rightarrow \alpha L= n\pi \Rightarrow \alpha={n\pi\over L} \Rightarrow X =c_2 \sin {n\pi x\over L}, \lambda=-{n^2\pi^2 \over L^2} \\ \Rightarrow T''+\alpha^2c^2 T=0 \Rightarrow T=c_3 \cos \alpha c t+ c_4 \sin \alpha ct =c_3\cos {n\pi c t\over L}+ c_4 \sin{n\pi ct \over L} \\ \Rightarrow u_n(x,t)=X(x)T(t) =c_2 \sin{n\pi x\over L}\left( c_3\cos {n\pi c t\over L}+ c_4 \sin{n\pi ct \over L} \right) \\\qquad =   \sin{n\pi x\over L}\left( A_n\cos {n\pi c t\over L}+ B_n \sin{n\pi ct \over L} \right) \\ \Rightarrow u(x,t) = \sum_{n=1}^\infty \sin{n\pi x\over L}\left( A_n\cos {n\pi c t\over L}+ B_n \sin{n\pi ct \over L} \right) \\ \Rightarrow {\partial \over \partial t}u(x,t)= \sum_{n=1}^\infty \sin{n\pi x\over L}\left( -A_n {n\pi c\over L}\sin {n\pi c t\over L}+ B_n{n\pi c\over L} \cos{n\pi ct \over L} \right)\\ \Rightarrow u(x,0)= f(x) =\sum_{n=1}^\infty A_n \sin{n\pi x\over L} \Rightarrow A_n= {2\over L} \int_0^L f(x) \sin{n\pi x\over L}\,dx\\ \Rightarrow  {\partial \over \partial t}u(x,0)=  g(x)= \sum_{n=1}^\infty B_n{n\pi c\over L} \sin{n\pi x\over L} \Rightarrow B_n ={2\over n\pi c} \int_0^L g(x) \sin{n\pi x\over L}\,dx \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty \sin{n\pi x\over L} \left( \left({2\over L} \int_0^L f(x) \sin{n\pi x\over L}\,dx \right) \cos{n\pi ct \over L}+  \left({2\over n\pi c} \int_0^L g(x) \sin{n\pi x\over L}\,dx \right) \sin{n\pi c t\over L} \right)\\ 故選\bbox[red, 2pt]{(D)}$$

解答: $$\nabla \cdot \mathbf F =\frac{\partial }{\partial x}F_1+ \frac{\partial }{\partial y}F_2+ \frac{\partial }{\partial z}F_3=0+0+0=0,故選\bbox[red, 2pt]{(A)}$$

解答: $$\nabla \times \mathbf F =\begin{vmatrix} \vec i& \vec j& \vec k\\ \frac{ \partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\F_1 & F_2 & F_3\end{vmatrix} = \begin{vmatrix} \vec i& \vec j& \vec k\\ \frac{ \partial }{\partial x} &\frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\y^2z & xz^2 & xy^2\end{vmatrix} \\=2xy \vec i+y^2\vec j+z^2\vec k-2yz\vec k-y^2\vec j-2xz\vec i =(2xy-2xz)\vec i+0\vec j+(z^2-2yz)\vec k\\=(2xy-2xz, 0,z^2-2yz),故選 \bbox[red, 2pt]{(F)}$$

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解題僅供參考，碩士班歷年試題及詳解