114年雲科大電機碩士班-工程數學詳解

 國立雲林科技大學114學年度碩士班招生考試

系所:電機系
科目:工程數學

解答:$$\textbf{A. }(y^2+3)dx =y\sec^2 x\,dy \Rightarrow {1\over \sec^2 x}dx={y\over y^2+3}\,dy \Rightarrow  \int \cos^2 x\,dx= \int{y\over y^2+3}\,dy \\ \quad \Rightarrow \int{1\over 2}(\cos 2x+1)\,dx =\int{y\over y^2+3}\,dy \Rightarrow {1\over 4}\sin 2x+{1\over 2}x +c_1={1\over 2}\ln(y^2+3) \\ \quad \Rightarrow {1\over 2}\sin 2x+x+2c_1 =\ln(y^2+3) \Rightarrow y^2=c_2e^{{1\over 2}\sin 2x+x}-3 \Rightarrow \bbox[red, 2pt]{y= \pm \sqrt{c_2e^{{1\over 2}\sin 2x+x}-3}} \\\textbf{B. }y'-y=4e^x \Rightarrow e^{-x}y'-e^{-x}y=4 \Rightarrow \left(e^{-x}y \right)'=4 \Rightarrow e^{-x}y=4x+c_1 \Rightarrow \bbox[red, 2pt]{y=4xe^x +c_1e^x} \\\textbf{C. }y'=10 \cos 5x \Rightarrow y=\int 10\cos 5x\,dx \Rightarrow \bbox[red, 2pt]{y=2\sin 5x+c_1}$$

解答:$$L\{y''\}-2 L\{ y'\}+ L\{y\} =3L\{t^2\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+ Y(s)=3\cdot {2\over s^3} \\ \Rightarrow (s^2-2s+1)Y(s)-1={6\over s^3 }\Rightarrow Y(s)={6\over s^3(s^2-2s+1) }+{1 \over s^2-2s+1} \\= \left(-{18\over s-1}+{6\over (s-1)^2}+{18\over s}+{12\over s^2}+{6\over s^3} \right) +{1\over (s-1)^2} =-{18\over s-1}+{7\over (s-1)^2}+{18\over s}+{12\over s^2}+{6\over s^3} \\ \Rightarrow y(t)=L^{-1}\left(-{18\over s-1}+{7\over (s-1)^2}+{18\over s}+{12\over s^2}+{6\over s^3} \right) =-18e^t+7te^t+18+12t+3t^2\\ \Rightarrow \bbox[red,2pt]{y(t)=-18e^t+7te^t+18+12t+3t^2}$$

解答:$$\textbf{A. }F(s)= L\{t^2\}+ L\{\cos^2 t\}+ L\{e^{-2t-5}\} ={2\over s^3}+ {1\over 2}L\{\cos 2t+1\} +e^{-5} L\{ e^{-2t}\} \\\qquad ={2\over s^2}+{1\over 2}\left({s\over s^2+2^2}+{1\over s} \right) +e^{-5}{1\over s+2} \Rightarrow \bbox[red, 2pt]{F(s)={2\over s^2}+{1\over 2s}+ {s\over 2(s^2+4)} +{1\over (s+2)e^5}} \\\textbf{B. }f(t) =L^{-1} \left[{s\over s^2+4} e^{-\pi s/2}\right] =u(t-{\pi\over 2}) L^{-1}\left[{s\over s^2+4}  \right](t-{\pi\over 2})  =u(t-{\pi\over 2})  \cos2(t-{\pi\over 2})\\ \quad \Rightarrow \bbox[red, 2pt]{f(t)=u(t-{\pi\over 2})  \cos(2t-\pi)} \\\textbf{C. }L\{f(t)\}=L\{2t^2\}-L\{e^{-2t}\}-L\left\{\int_0^t f(\tau) e^{t-\tau}\,d \tau \right\} \Rightarrow F(s)= {4\over s^3}-{1\over s+2}-F(s) \cdot {1\over s-1} \\ \quad \Rightarrow {s\over s-1}F(s)= {4\over s^3}-{1\over s+2} \Rightarrow F(s)= {4(s-1)\over s^4}-{s-1\over s(s+2)} ={4\over s^3}-{4\over s^4}+{1\over 2s}-{3\over 2(s+2)} \\ \Rightarrow f(t)=L^{-1}\{F(s)\}\Rightarrow \bbox[red, 2pt]{y(t) =2t^2-{2\over 3}t^3+{1\over 2}-{3\over 2}e^{-2t}}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow m(m-1)x^m+2mx^m+x^m=0 \\ \Rightarrow (m^2+m+1)x^m=0 \Rightarrow m^2+m+1=0 \Rightarrow m={-1\pm \sqrt 3i\over 2} \\\Rightarrow y_h= {1\over \sqrt x}(c_1 \cos ({\sqrt 3\over 2}\ln x) +c_2\sin ({\sqrt 3\over 2}\ln x))\\ y_p=ax^4+bx^3+cx^2+dx +e \Rightarrow y_p'=4ax^3+3bx^2+2cx+d \Rightarrow y_p''=12ax^2+6bx+2c \\ \Rightarrow x^2y_p''+2xy_p'+y_p'=21 ax^4+13bx^3+7cx^2+3dx+e =3x^4 \Rightarrow \cases{a=1/7\\ b=c=0\\ d=e=0} \\ \Rightarrow y_p={1\over 7}x^4 \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y={1\over \sqrt x}(c_1 \cos ({\sqrt 3\over 2}\ln x) +c_2\sin ({\sqrt 3\over 2}\ln x))+{1\over 7}x^4}$$

解答:$$\cases{x+y+z=3\\ 4x+2y+3z=1\\ 2x-y+z=2} \Rightarrow \begin{bmatrix} 1& 1& 1\\ 4& 2& 3\\ 2&-1& 1\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix}= \begin{bmatrix}3\\ 1\\ 2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\text{augmented matrix: } \begin{bmatrix} 1& 1& 1& 3\\ 4& 2& 3& 1\\ 2&-1& 1& 2\end{bmatrix} }\\ \begin{bmatrix} 1& 1& 1& 3\\ 4& 2& 3& 1\\ 2&-1& 1& 2\end{bmatrix} \xrightarrow{R_2-4R_1\to R_2, R_3-2R_1\to R_3} \begin{bmatrix} 1 & 1 & 1 & 3\\0 & -2 & -1 & -11\\0 & -3 & -1 & -4 \end{bmatrix} \xrightarrow{R_2/(-2)\to R_2} \begin{bmatrix} 1 & 1 & 1 & 3\\0 & 1 & \frac{1}{2} & \frac{11}{2}\\0 & -3 & -1 & -4 \end{bmatrix}  \\\xrightarrow{R_1 -R_2\to R_1, R_3+3R_2\to R_3} \begin{bmatrix} 1 & 0 & \frac{1}{2} & - \frac{5}{2}\\0 & 1 & \frac{1}{2} & \frac{11}{2}\\0 & 0 & \frac{1}{2} & \frac{25}{2} \end{bmatrix} \xrightarrow{2R_3 \to R_3} \begin{bmatrix} 1 & 0 & \frac{1}{2} & - \frac{5}{2}\\0 & 1 & \frac{1}{2} & \frac{11}{2}\\0 & 0 & 1 & 25 \end{bmatrix} \\ \xrightarrow{R_1-R_3/2\to R_1, R_2-R_3/2\to R_2 } \begin{bmatrix} 1 & 0 & 0 & -15\\0 & 1 & 0 & -7\\0 & 0 & 1 & 25 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{x=-15\\ y=-7\\ z=25}}$$

解答:$$A= \begin{bmatrix} 4 & 1 & 1 \\0 & 3 & -1 \\0 & 0 & 5\end{bmatrix}  \Rightarrow \det(A-\lambda I)= -(\lambda-4)(\lambda-3)(\lambda-5)=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: 3,4,5}} \\ \lambda_1=3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 1 & 1 & 1 \\0 & 0 & -1 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1+x_2=0\\ x_3=0} \\\qquad \Rightarrow v= \begin{pmatrix} -x_2\\x_2\\0 \end{pmatrix}, 取v_1= \begin{pmatrix} -1\\1\\0 \end{pmatrix} \\ \lambda_2=4 \Rightarrow (A-\lambda_2 I)v= 0 \Rightarrow \begin{bmatrix} 0 & 1 & 1 \\0 & -1 & -1 \\0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{ x_2=0\\ x_3=0} \\\qquad \Rightarrow v= \begin{pmatrix} x_1\\0\\0 \end{pmatrix}, 取v_2 = \begin{pmatrix} 1\\0 \\0 \end{pmatrix} \\ \lambda_3=5 \Rightarrow (A-\lambda_3 I)v =0 \Rightarrow \begin{bmatrix} -1 & 1 & 1 \\0 & -2 & -1 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{ 2x_1=x_3\\ 2x_2 +x_3=0} \\\qquad \Rightarrow v= \begin{pmatrix} x_3/2\\-x_3/2 \\x_3 \end{pmatrix}, 取v_3 =\begin{pmatrix} 1\\-1 \\2 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: }\begin{pmatrix} -1\\1\\0 \end{pmatrix}, \begin{pmatrix} 1\\0 \\0 \end{pmatrix}, \begin{pmatrix} 1\\-1 \\2 \end{pmatrix}}$$

解答:$$f(x)=\pi^2+x^2 \Rightarrow f(-x)=f(x) \Rightarrow f(x)\text{ is even} \Rightarrow b_n=0\\ a_0={1\over 2\pi }\int_{-\pi}^\pi (\pi^2+x^2) \,dx ={1\over 2\pi}\cdot {8\over 3}\pi^3 ={4\over 3}\pi^2 \\a_n={1\over \pi }\int_{-\pi}^\pi (\pi^2+x^2) \cos(nx)\,dx ={1\over \pi}\cdot {4\pi \over n^2}(-1)^n ={4\over n^2}(-1)^n \\ \Rightarrow f(x)=a_0+ \sum_{n=1}^\infty a_n \cos(nx) =\bbox[red, 2pt]{{4\over 3}\pi^2+ \sum_{n=1}^\infty {4\over n^2}(-1)^n \cos(nx)}$$

解答:$$A=\begin{bmatrix}3& 2& 4\\ 2& 0 & 2\\ 4& 2& 3 \end{bmatrix} \xrightarrow{R_2-({\color{blue}{2/3}})R_1 \to R_2, R_3-({\color{blue}4/3})R_1 \to R_3} \begin{bmatrix}3 & 2 & 4\\0 & - \frac{4}{3} & - \frac{2}{3}\\0 & - \frac{2}{3} & - \frac{7}{3} \end{bmatrix} \xrightarrow{R_3-({\color{blue}1/2})R_2\to R_3} \begin{bmatrix}3 & 2 & 4\\0 & - \frac{4}{3} & - \frac{2}{3}\\0 & 0 & -2 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{L=\begin{bmatrix}1 & 0 & 0\\{\color{blue}{2 \over 3}} & 1 & 0\\{\color{blue}{4\over 3}} & {\color{ blue}{1\over 2} }& 1 \end{bmatrix}, U=\begin{bmatrix}3 & 2 & 4\\0 & - \frac{4}{3} & - \frac{2}{3}\\0 & 0 & -2 \end{bmatrix}}$$
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