111年成大資源工程碩士班-工程數學詳解

 國立成功大學111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: $$\textbf{(a)}\; y''+y=0 \Rightarrow y_h=c_1\cos x+c_2\sin x\\ y_p=A\cos(2x)+ B\sin(2x) \Rightarrow y_p'=-2A\sin(2x)+2 B\cos(2x) \Rightarrow y_p''=-4A\cos(2x)-4B\sin(2x) \\ \Rightarrow y_p''+y_p=-3A\cos(2x)-3B\sin(2x)=\sin(2x) \Rightarrow \cases{A=0 \\B=-1/3} \Rightarrow y_p=-{1\over 3}\sin(2x) \\ \Rightarrow y=y_h+y_p \Rightarrow y=c_1\cos x+c_2\sin x-{1\over 3}\sin(2x) \Rightarrow y'=-c_1\sin x+c_2\cos x-{2\over 3}\cos(2x) \\ \Rightarrow \cases{y(0)=c_1=0\\ y'(0)=c_2-2/3=0} \Rightarrow c_2={2\over 3} \Rightarrow \bbox[red, 2pt]{y={2\over 3}\sin x-{2\over 3} \sin(2x)} \\\textbf{(b)}\; y'=y+e^x \Rightarrow y'-y=e^x \Rightarrow \text{ integration factor }I(x)=e^{\int -1\,dx} =e^{-x} \Rightarrow e^{-x}y'-e^{-x}y=1 \\\quad \Rightarrow (e^{-x}y)'=1 \Rightarrow e^{-x}y=x+c_1 \Rightarrow y=xe^x+ c_1e^x \Rightarrow y(0)=c_1=2 \Rightarrow \bbox[red, 2pt]{y=xe^x +2e^x}$$

解答: $$\textbf{(a)}\; \int_0^\infty e^{3t+2}e^{-st}\,dt = e^2\int_0^\infty e^{(3-s)t}\,dt =e^2\left. \left[ {1\over 3-s}e^{(3-s)t}\right] \right|_0^\infty = \bbox[red, 2pt]{e^2\over s-3} \\\textbf{(b)}\; L^{-1}\left\{ {s+6\over s^2+4s+20}\right\} =L^{-1}\left\{ {(s+2)+ 4\over (s+2)^2+4^2}\right\} = \bbox[red, 2pt]{e^{-2t}(\cos 4t+ \sin 4t)}$$

解答: $$\textbf{(a-1)}\left[\begin{matrix}1 & 3 & 4 & 2\\2 & -1 & 1 & -3\\1 & 2 & 3 & 1\\1 & 3 & 0 & 1\end{matrix}\right] \xrightarrow{R_2-2R_1\to R_2, R_3-R_1\to R_3, R_4-R_1\to R_4} \left[\begin{matrix}1 & 3 & 4 & 2\\0 & -7 & -7 & -7\\0 & -1 & -1 & -1\\0 & 0 & -4 & -1\end{matrix} \right] \xrightarrow{R_1+3R_3\to R_1, R_2-7_R3\to R_2} \\\left[\begin{matrix}1 & 0 & 1 & -1\\0 & 0 & 0 & 0\\0 & -1 & -1 & -1\\0 & 0 & -4 & -1\end{matrix}\right] \xrightarrow{-R_3\to R_3, -R_4/4 \to R_4} \left[\begin{matrix}1 & 0 & 1 & -1\\0 & 0 & 0 & 0\\0 & 1 & 1 & 1\\0 & 0 & 1 & 1/4\end{matrix}\right] \xrightarrow{R_1-R_4\to R_1,R_2 \leftrightarrow R_3}  \left[ \begin{matrix}1 & 0 & 0 & -5/4\\0 & 1 & 1 & 1\\0 & 0 & 0 & 0\\0 & 0 & 1 & 1/4\end{matrix} \right] \\ \xrightarrow{R_2-R_4\to R_2, R_3\leftrightarrow R_4} \bbox[red,2pt]{\left[ \begin{matrix}1 & 0 & 0 & -5/4\\0 & 1 & 0 & 3/4\\0 & 0 & 1 & 1/4 \\0 & 0 & 0 & 0\end{matrix} \right] } \\\textbf{(a-2)} \; 由rref(A) 可知rank(A)＝\bbox[red, 2pt]3 \\\textbf{(a-3)} \;\text{rank}(A)=1-1+3+1=\bbox[red, 2pt]4 \\\textbf{(b)} \;B=\begin{bmatrix} 0 & 1 & 0 \\1 & 0 & 0 \\0 & 0 & 1 \end{bmatrix} \Rightarrow \det(B-\lambda I)=-(\lambda-1)^2(\lambda+1)=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }1,-1} \\ \lambda_1=1 \Rightarrow (B-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} -1 & 1 & 0 \\1 & -1 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow x_1=x_2\\\qquad  \Rightarrow v=x_1 \begin{pmatrix}1\\ 1\\ 0 \end{pmatrix}+ x_3\begin{pmatrix}0 \\ 0\\ 1 \end{pmatrix} , 取v_1=\begin{pmatrix}1\\ 1\\ 0 \end{pmatrix}, v_2 = \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}\\ \lambda_2=-1 \Rightarrow (B-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} 1 & 1 & 0 \\1 & 1 & 0 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow  \cases{x_1+x_2=0\\ x_3=0} \\ \qquad \Rightarrow v=x_2 \begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix} , 取v_3=\begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: } \begin{pmatrix}1\\ 1\\ 0 \end{pmatrix},  \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}, \begin{pmatrix}-1\\ 1\\ 0 \end{pmatrix}}$$

解答: $$\textbf{(a)} \; \vec F(x,y,z)=(3xy^2,-2yz^2, 4x^2z) \Rightarrow \text{div }\vec F=\frac{\partial }{\partial x}3xy^2 +\frac{\partial  }{\partial y}(-2yz^2) +\frac{\partial  }{\partial z} 4x^2z \\\quad =3y^2-2z^2+4x^2  \Rightarrow \text{div }\vec F(-1,2,3)=12-18+4= \bbox[red, 2pt]{-2} \\\textbf{(b)} \; \cases{F=x+y^2+z^3\\ \vec v=(2,2,1)} \Rightarrow \cases{(F_x, F_y,F_z)= (1,2y,3z^2)\\ \vec n={\vec v\over \Vert \vec v\Vert} =({2\over 3}, {2\over 3}, {1\over 3})}  \Rightarrow D_{\vec n}F(x,y,z) =(1,2y,3z^2) \cdot ({2\over 3}, {2\over 3}, {1\over 3})\\\quad  ={2\over 3}+{4y\over 3}+z^2 \Rightarrow D_{\vec n}F(3,2,-1) ={2\over 3}+{8\over 3}+1 = \bbox[red, 2pt]{13\over 3}\\ \textbf{(c)} \; 假設C為平滑曲線所組合而成的封閉曲線, 所圍區域為R,\\而P(x,y)及Q(x,y)為連續且一次偏導數也連續,則\\ \oint_C Pdx+Qdy = \iint_R {\partial Q\over \partial x} -{\partial P\over \partial y}dA$$

解答: $$\textbf{(a)} \; f(x)為奇函數 \Rightarrow a_n=0,而b_n= \int_{-1}^0 -\sin(n\pi x)\,dx + \int_0^1 \sin(n\pi x)\,dx ={2\over n\pi}(1-(-1)^n)\\\quad  \Rightarrow f(x)= \bbox[red, 2pt]{\sum_{n=1}^\infty {2\over n\pi}(1-(-1)^n) \sin(n\pi x)} \\\textbf{(a)} \; f(x)=e^{-x}, x\gt 0 \Rightarrow A(\alpha)=\int_0^\infty e^{-x} \cos(\alpha x)\,dx ={1\over 1+\alpha^2} \\ \quad \Rightarrow \bbox[red, 2pt]{ f(x)={2\over \pi}\int_0^\infty {\cos (\alpha x)\over 1+\alpha^2} \,dx} \\\textbf{(c)} \; F(\omega) =\int_{-\infty}^\infty e^{-|x|} e^{-j\omega x}\,dx =2\int_0^\infty e^{-x} e^{-j\omega x}\,dx = 2\int_0^\infty   e^{-(j\omega+1) x}\,dx =2 \left. \left[ -{1\over j\omega +1} e^{-(j\omega+1)x}\right] \right|_0^\infty\\ \qquad =\bbox[red, 2pt]{2\over j\omega +1}$$

解答: $$\text{for example, RLC circuit}$$

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

解題僅供參考,其他歷年試題及詳解