113年成大資源工程碩士班-工程數學詳解

 國立成功大學113學年度碩士班招生考試

系所:資源工程學系
科目:工程數學

解答: $$\textbf{(a)}\;{dy\over dx}=y+x^2 \Rightarrow y'-y=x^2 \Rightarrow e^{-x}y'-ye^{-x}=x^2e^{-x} \Rightarrow (e^{-x}y)'=x^2e^{-x} \\ \Rightarrow e^{-x}y= \int x^2 e^{-x}\,dx =-e^{-x}(x^2+2x+2)+c_1 \Rightarrow y=-(x^2+2x+2)+ c_1e^x\\ \Rightarrow y(0)=-2+c_1=1 \Rightarrow c_1=3 \Rightarrow \bbox[red, 2pt]{y=3e^x-(x^2+2x+2)} \\\textbf{(b)}\;y''+y=0 \Rightarrow y_h=c_1\cos x+c_2\sin x \\ \Rightarrow y_p= Ax\cos x+Bx\sin x \Rightarrow y_p'=A\cos x+B\sin x-Ax\sin x+Bx\cos x\\ \Rightarrow y_p''=-2A\sin x+2B\cos x -Ax\cos x -Bx\sin x \\ \Rightarrow y_p''+y_p=-2A\sin x+2B\cos x =\cos(x) \Rightarrow \cases{A=0\\ B=1/2} \Rightarrow y_p={1\over 2}x\sin x\\ \Rightarrow y=y_h+ y_p = c_1\cos x+c_2\sin x +{1\over 2}x\sin x \Rightarrow y'=({1\over 2}-c_1)\sin x+c_2\cos x+ {1\over 2}x\cos x \\ \Rightarrow \cases{y(0)= c_1= 0\\ y'(0)=c_2=0} \Rightarrow \bbox[red, 2pt]{y={1\over 2}x\sin x}$$

解答: $$\textbf{(a)}\;L\{\exp(3t-2)\} =\int_0^\infty e^{3t-2}e^{-st}\,dt  = e^{-2}\int_0^\infty e^{(3-s)t} \,dt  =e^{-2}\left. \left[ {1\over 3-s}e^{(3-s)t} \right] \right|_0^\infty =\bbox[red, 2pt]{e^{-2}\over s-3} \\\textbf{(b)}\; L^{-1} \left\{{s+2\over (s+1)s} \right\} = L^{-1} \left\{{2\over s}-{1\over s+1} \right\}= L^{-1} \left\{{2\over s}  \right\}- L^{-1} \left\{{1\over s+1} \right\} =\bbox[red, 2pt]{2-e^{-t}}$$

解答: $$\textbf{(a-1)}\;A=\begin{bmatrix}1 & 1 & 1 & 2\\1 & 1 & 2 & 1\\1 & 2 & 1 & 1\\2 & 1 & 1 & 1 \end{bmatrix} \xrightarrow{R_2-R_1\to R_2, R_3-R_1\to R_3,R_4-2R_1\to R_4} \begin{bmatrix}1 & 1 & 1 & 2\\0 & 0 & 1 & -1\\0 & 1 & 0 & -1\\0 & -1 & -1 & -3 \end{bmatrix} \\ \xrightarrow{R_1-R_3\to R_1,R_1+R_4\to R_4} \begin{bmatrix}1 & 0 & 1 & 3\\0 & 0 & 1 & -1\\0 & 1 & 0 & -1\\0 & 0 & -1 & -4 \end{bmatrix} \xrightarrow{R_4+R_2\to R_4} \begin{bmatrix}1 & 0 & 1 & 3\\0 & 0 & 1 & -1\\0 & 1 & 0 & -1\\0 & 0 & 0 & -5 \end{bmatrix} \\ \xrightarrow{ R_2 \leftrightarrow R_3, R_4=R_4/(-5)} \begin{bmatrix}1 & 0 & 1 & 3\\0 & 1 & 0 & -1\\0 & 0 & 1 & -1\\0 & 0 & 0 & 1 \end{bmatrix} \xrightarrow{R_1-R_3\to R_1,R_2+R_4}{\begin{bmatrix}1 & 0 & 0 & 4\\0 & 1 & 0 & 0\\0 & 0 & 1 & -1\\0 & 0 & 0 & 1 \end{bmatrix}} \\ \xrightarrow{R_1-4R_4\to R_1,R_3+R_4\to R_3} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix} =I_4 \Rightarrow rref(A)=\bbox[red, 2pt]{\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1 \end{bmatrix}}\\\textbf{(a-2)}\; rref(A)=I_4 \Rightarrow rank(A)=4 \Rightarrow nullity(A)=4-rank(A)= \bbox[red, 2pt]0\\\textbf{(b)}\;  B=\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} \Rightarrow \det(B-\lambda I)=0 \Rightarrow \lambda^2-2\lambda =0 \Rightarrow \lambda=0,2 \\ \lambda_1=0 \Rightarrow (B-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} =0   \Rightarrow x_1+ x_2=0 \Rightarrow v= x_2\begin{pmatrix} -1 \\1 \end{pmatrix},取v_1=\begin{pmatrix}-1 \\1 \end{pmatrix} \\\lambda_2=2 \Rightarrow (B-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -1 & 1 \\1 & -1 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} =0   \Rightarrow x_1=x_2 \Rightarrow v=x_2 \begin{pmatrix}1 \\1 \end{pmatrix},取v_2= \begin{pmatrix}1 \\1 \end{pmatrix} \\ 因此 \text{ eigenvalues: }\bbox[red, 2pt]{0,2 }\text{ and the corresponding eigenvectors: } \bbox[red, 2pt]{\begin{pmatrix}-1 \\1 \end{pmatrix},\begin{pmatrix}1 \\1 \end{pmatrix}}$$

解答: $$\textbf{(a)}\;\vec F=3x^2y^3\vec i-2y^3z^2\vec j+x^3z^2\vec k \Rightarrow \text{curl }\vec F=\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\\vec F_1 & \vec F_2 & \vec F_3\end{vmatrix} =\begin{vmatrix}\vec i & \vec j& \vec k \\{\partial \over \partial x} & {\partial \over \partial y} &{\partial \over \partial z} \\3x^2y^3 & -2y^3z^2 & x^3z^2\end{vmatrix} \\ =0\vec i+ 0\vec j+0\vec k -9x^2y^2\vec k-3x^2z^2 \vec j+4y^3z \vec i  \Rightarrow \text{curl }\vec F(2,-1,1) =\bbox[red, 2pt]{-4\vec i-12\vec j-36\vec k} \\\textbf{(b)}\; \cases{f(x,y, z)={xy^2\over z^3} \Rightarrow \nabla f =({y^2\over z^3},{2xy\over z^3}, -{3xy^2\over z^4} ) \\ \vec n= 2\vec i+2\vec j+ 1\vec k \Rightarrow{\vec n\over |\vec n|}={2\over 3}\vec i+ {2\over 3}\vec j+{1\over 3}\vec k}   \\ \Rightarrow \nabla f\cdot {\vec n\over |\vec n|} ={2\over3}\cdot {y^2\over z^3} +{2\over 3}\cdot {2xy\over z^3 } -{1\over3}\cdot {3xy^2\over z^4}\\ \Rightarrow \nabla f(3,2,1)\cdot {\vec n\over |\vec n|}  ={8\over 3}+8-12 = \bbox[red, 2pt]{-{4 \over 3}}$$

$$\textbf{(c)}\;\text{Let }M(x,y) \text{ and }N(x,y) \text{ be continuous and have continuous partial derivatives } \\\text{in a region }R \text{ and on its boundary }C. \text{ Then, for both simply-connected and }\\\text{ multiply-connected }R.\\ \oint_C (Mdx+ Ndy)= \displaystyle \iint_R \left( {\partial N\over \partial x} -{\partial M\over \partial y}\right)\,dxdy$$

(d) Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Also let $\vec F$ be a vector field then, $$\int_C \vec F\cdot d\vec r= \iint_S \text{curl }\vec F\cdot d\vec S$$

解答: $$\textbf{(a)}\;a_0= {1\over 2}\int_{-1}^1 f(x)  \,dx = {1\over 2}\int_{0}^1   1\,dx  ={1\over 2} \\ a_n= \int_0^1\cos(n\pi x)\,dx =\left.\left[{1\over n\pi} \sin(n\pi x) \right] \right|_0^1=0\\ b_n= \int_0^1 \sin(n\pi x)\,dx =\left.\left[-{1\over n\pi} \cos(n\pi x) \right] \right|_0^1= {1\over n\pi}(1-(-1)^n) \\ \Rightarrow f(x)=\bbox[red, 2pt]{{1\over 2}+\sum_{n=1}^\infty  {1\over n\pi}(1-(-1)^n) \sin(n\pi x)} \\ \textbf{(b)}\;\mathcal F(\omega) =\int_{-\infty}^\infty f(t)e^{-j\omega t}\,dt =\int_0^\infty ke^{-at} e^{-j\omega t}\,dt = k\int_0^\infty   e^{-(a+j\omega) t}\,dt =k \left. \left[-{1\over a+j\omega} e^{-(a+j\omega) t}\right] \right|_0^\infty \\= \bbox[red, 2pt]{k\over a+j\omega}$$

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