114年中山大學海工碩士班-工程數學詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：工程數學【離岸風電碩士班碩士班選考、海工系碩士班選考、海工聯合碩士班選考】

解答: $$\lambda^2-4\lambda+4=(\lambda-2)^2 =0 \Rightarrow \lambda=2 \Rightarrow y=c_1e^{2x} +c_2 xe^{2x} \Rightarrow y'=(2c_1 +c_2)e^{2x}+  2c_2xe^{2x} \\ \Rightarrow \cases{y(0)= c_1=3\\ y'(0)=2c_1+c_2=1} \Rightarrow \cases{c_1=3\\ c_2=-5} \Rightarrow \bbox[red, 2pt]{y=3e^{2x}-5xe^{2x}}$$

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

解答: $$\left[ \begin{array}{rrr|rrr} -1& 1& 2& 1& 0 & 0\\ 3& -1& 1& 0& 1& 0\\ -1& 3& 4& 0& 0& 1\end{array} \right] \xrightarrow {3R_1+R_2 \to R_2, R_3-R_1\to R_3} \left[ \begin{array}{rrr|rrr}-1 & 1 & 2 & 1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 2 & 2 & -1 & 0 & 1\end{array} \right] \xrightarrow{-R_1\to R_1,R_3-R_2\to R_3} \\ \left[ \begin{array}{rrr|rrr}1 & -1 & -2 & -1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 0 & -5 & -4 & -1 & 1\end{array} \right] \xrightarrow{R_2/2\to R_2, R_3/(-5) \to R_3} \left[ \begin{array}{rrr|rrr}1 & -1 & -2 & -1 & 0 & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \\ \xrightarrow{R_1 +R_2\to R_2}  \left[ \begin{array}{rrr|rrr}1 & 0 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \xrightarrow{R_1-(3/2)R_3\to R_1, R_2-(7/2) R_3\to R_2}  \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \\ \Rightarrow \left[ \begin{array}{rrr|rrr} -1& 1& 2 \\ 3& -1& 1 \\ -1& 3& 4 \end{array} \right] ^{-1} = \bbox[red, 2pt] {\left[ \begin{array}{rrr|rrr} - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\  - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\  \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right]}$$

解答: $$\textbf{(1) }\text{div }xyz(x\mathbf i+y\mathbf j+z\mathbf k) =\text{div }(x^2yz \mathbf i+xy^2z \mathbf j+ xyz^2\mathbf k) \\\qquad ={\partial \over \partial x}(x^2yz) +{\partial \over \partial y}(xy^2z) +{\partial \over \partial z}(xyz^2) =2xyz+2xyz+2xyz= \bbox[red, 2pt]{6xyz}\\ \textbf{(2) }\text{curl }xyz(x\mathbf i+y\mathbf j+z\mathbf k) =\text{curl }(x^2yz \mathbf i+xy^2z \mathbf j+ xyz^2\mathbf k) \\\qquad =\begin{vmatrix}\mathbf i&  \mathbf j & \mathbf k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z}  \\ x^2yz& xy^2z & xyz^2\end{vmatrix}= \bbox[red, 2pt]{(xz^2-xy^2) \mathbf i+(x^2y-yz^2) \mathbf j +(y^2z-x^2z) \mathbf k}$$

解答: $$L\{e^{at}\} =\int_0^\infty e^{at}\cdot e^{-st}\,dt = \int_0^\infty e^{-(s-a)t} \,dt = \left. \left[ -{1\over s-a}e^{-(s-a)t} \right] \right|_0^\infty =\bbox[red, 2pt]{1\over s-a}$$

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

解答: $$f(-x)=-f(x) \Rightarrow f(x) \text{ is odd} \Rightarrow a_n=0\\ b_n={1\over \pi} \int_{-\pi}^\pi f(x) \sin(nx)\,dx ={2\over \pi}\int_0^\pi k\sin(nx)\,dx ={2k\over n\pi}(1-(-1)^n) \\ \Rightarrow \bbox[red, 2pt]{f(x)=\sum_{n=1}^\infty {2k\over n\pi}(1-(-1)^n) \sin(nx)\,dx}$$

解答: $$\text{Suppose }u(x,t)=X(x)T(t), \text{ then} {\partial^2u \over \partial t^2} =c^2{\partial^2 u\over \partial x^2}  \Rightarrow XT'' = c^2X''T \Rightarrow {T''\over c^2T} ={X''\over X}= \alpha\\ \text{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 }\alpha=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow B.C.: \cases{X(0)=c_2=0\\ X(L)=c_1L+c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0}\\\qquad \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case II } \alpha=\rho^2 \gt 0 \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1e^{\rho x}+ c_2 e^{-\rho x} \\\qquad \Rightarrow \text{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 \Rightarrow u=0\\ \textbf{Case III }\alpha= -\rho^2 \lt 0 \Rightarrow X''+\rho^2 X=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 \Rightarrow \rho={n\pi\over L}\\ \qquad \Rightarrow X_n= c_2\sin({n\pi x\over L}),n=1,2,\dots \\\Rightarrow T''+ \rho^2c^2 T=0 \Rightarrow T=c_3\cos(\rho c t)+ c_4\sin(\rho c t) \Rightarrow T_n=c_3 \cos({nc \pi t\over L}) + c_4\sin ({nc \pi t\over L}) \\ \Rightarrow u_n= c_2\sin({n\pi x\over L})\left( c_3 \cos({nc \pi t\over L}) + c_4\sin ({nc \pi t\over L}) \right) \\ \Rightarrow u(x,t)= \sum_{n= 1}^\infty \sin({n\pi x\over L})\left( a_n \cos({nc \pi t\over L}) + b_n\sin ({nc \pi t\over L}) \right)  \\ \Rightarrow u_t(x,t)=\sum_{n=1}^\infty \sin({n\pi x\over L}) \left(-a_n{nc\pi\over L}\sin {nc\pi t\over L}+ b_n{nc\pi\over L} \cos{nc\pi t\over L} \right)\\\text{I.C.: } u(x,0)= \sum_{n=1}^\infty a_n\sin({n\pi x\over L})     =\sin{\pi x\over L} \Rightarrow \begin{cases}a_1=1\\ a_n=0, & n\ne 1 \end{cases} \\ \text{I.C.: } u_t(x,0)=\sum_{n=1}^\infty   b_n{nc\pi\over L}\sin({n\pi x\over L})  =0 \Rightarrow b_n=0 \\ \Rightarrow \bbox[red, 2pt] {u(x,t)=\sin{\pi x\over L}\cos {c\pi t\over L}}$$

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