113年雲科大電機碩士班-工程數學詳解

國立雲林科技大學113學年度碩士班招生考試

系所：電機系
科目：工程數學

解答： $$\textbf{(1)}\;\frac{d y}{dx}=e^{2x-3y} \Rightarrow e^{3y}dy=e^{2x}dx \Rightarrow {1\over 3}e^{3y} ={1\over 2} e^{2x}+c_1 \Rightarrow e^{3y} ={3\over 2}e^{2x}+c_2 \\\quad \Rightarrow 3y= \ln \left({3\over 2}e^{2x}+c_2 \right) \Rightarrow \bbox[red, 2pt] {y={1\over 3}\ln \left({3\over 2}e^{2x}+c_2 \right)} \\\textbf{(2)}\; x\frac{d y}{dx} -y= x^2\sin x   \Rightarrow {1\over x}\frac{d y}{dx}-{1\over x^2}y= \sin x \Rightarrow \left( {y\over x}\right)'=\sin x \Rightarrow {y\over x}=-\cos x+c_1 \\\quad \Rightarrow \bbox[red, 2pt]{y=-x\cos x+c_1x} \\\textbf{(3)}\; y^3dx+3xy^2dy=0 \Rightarrow 3xy^2dy=-y^3dx \Rightarrow {1\over y}dy =-{1\over 3x}dx \Rightarrow \ln y=-{1\over 3}\ln x+c_1 \\\quad \Rightarrow \bbox[red, 2pt]{y={c_2\over \sqrt[3] x}}$$

解答： $$5y''+y'=0 \Rightarrow 5\lambda^2+\lambda=\lambda(5\lambda + 1) =0 \Rightarrow \lambda=0,-{1\over 5} \Rightarrow y_h=c_1+c_2 e^{-x/5} \\ y_p=Ax^2+Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \Rightarrow 5y_p''+y_p'= 2Ax+10A+B=-6x\\ \Rightarrow \cases{2A=-6\\ 10A+B=0} \Rightarrow \cases{A=-3\\ B=30 } \Rightarrow y_p=-3x^2+30x+C \Rightarrow y=y_h+y_p \\ \Rightarrow y=c_3+c_2e^{-x/5}-3x^2+30x \Rightarrow y'=-{1\over 5}c_2-6x+30 \Rightarrow \cases{y(0)=c_3+c_2=0\\ y'(0)=-{1\over 5}c_2+30=-10}\\ \Rightarrow \cases{c_3=-200\\ c_2=200} \Rightarrow \bbox[red, 2pt]{y=-200+200e^{-x/5}-3x^2+30x}$$

解答： $$\textbf{(1)}\; L\{e^t(\cos \omega t-2\sin \omega t)\} =L\{e^t\cos \omega t\} -2L\{e^t \sin \omega t\} ={s-1\over (s-1)^2+\omega^2}-2\cdot {\omega\over (s-1)^2+\omega^2} \\\quad = \bbox[red,2pt]{s-1-2\omega \over (s-1)^2+\omega^2} \\\textbf{(2)}\; L^{-1}\left\{ {3\over s(s^2+9)}\right\} =L^{-1}\left\{{1\over 3s}- {s\over 3(s^2+9)}\right\} = \bbox[red, 2pt]{{1\over 3}-{1\over 3}\cos(3t)} \\ \textbf{(3)}\; L\left\{ \int_0^t \tau e^{t-\tau} \,d\tau \right\} =L\{t \} L\{e^t\} ={1\over s^2} \cdot {1\over s-1} =\bbox[red, 2pt]{1\over s^2(s-1)}$$

解答： $$a_0={1\over 2\pi} \int_{-\pi}^\pi (\pi+t)\,dt=\pi \\ a_n={1\over \pi} \int_{-\pi}^\pi (\pi+t) \cos(nt)\,dt = {1\over \pi} \left( \left. \left[ {\pi \over n}\sin(nt) +{t\over n}\sin(nt)+{1\over n^2} \cos(nt)\right] \right|_{-\pi}^\pi\right)=0\\ b_n={1\over \pi} \int_{-\pi}^\pi (\pi+t) \sin(nt)\,dt  =-{2 \over n}(-1)^n \\ \Rightarrow \bbox[red, 2pt]{f(x)=\pi-\sum_{n=1}^\infty {2 \over n}(-1)^n \sin(nx)}$$

解答： $$\cases{P_0(2,3,-4) \\P_1(3,-2,5)} \Rightarrow \vec v=\overrightarrow{P_0P_1} =(1,-5,9) \Rightarrow L:{x-2\over 1} ={y-3\over -5}={z+4\over 9} \\ \Rightarrow \bbox[red, 2pt]{L=(t+2,-5t+3,9t-4), t\in \mathbb R}$$

解答： $$\textbf{(1)}\; [v_1 \mid v_2 \mid v_3 \mid w_1\mid w_2\mid w_3] =\begin{bmatrix}2 & 1 & 1 & 6 & 4 & 5\\0 & 2 & 1 & 3 & -1 & 5\\1 & 0 & 1 & 3 & 3 & 2 \end{bmatrix} \xrightarrow{R_1/2\to R_1,R_2/2\to R_2} \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & 3 & 2 & \frac{5}{2}\\0 & 1 & \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} & \frac{5}{2}\\1 & 0 & 1 & 3 & 3 & 2 \end{bmatrix} \\ \xrightarrow{R_3-R_1\to R_3} \begin{bmatrix}1 & \frac{1}{2} & \frac{1}{2} & 3 & 2 & \frac{5}{2}\\0 & 1 & \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} & \frac{5}{2}\\0 & - \frac{1}{2} & \frac{1}{2} & 0 & 1 & - \frac{1}{2} \end{bmatrix} \xrightarrow{R_1-0.5R_2\to R_1,R_3+0.5R_2\to R_3} \begin{bmatrix}1 & 0 & \frac{1}{4} & \frac{9}{4} & \frac{9}{4} & \frac{5}{4}\\0 & 1 & \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} & \frac{5}{2}\\0 & 0 & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} & \frac{3}{4} \end{bmatrix} \\ \xrightarrow{(4/ 3)R_3 \to R_3} \begin{bmatrix}1 & 0 & \frac{1}{4} & \frac{9}{4} & \frac{9}{4} & \frac{5}{4}\\0 & 1 & \frac{1}{2} & \frac{3}{2} & - \frac{1}{2} & \frac{5}{2}\\0 & 0 & 1 & 1 & 1 & 1 \end{bmatrix} \xrightarrow{R_1-(1/4)R_3\to R_1,R_2-(1/2)R_3 \to R_2} \begin{bmatrix}1 & 0 & 0 & 2 & 2 & 1\\0 & 1 & 0 & 1 & -1 & 2\\0 & 0 & 1 & 1 & 1 & 1 \end{bmatrix} \\ \Rightarrow P_{S\leftarrow T} = \bbox[red, 2pt]{\left[\begin{matrix}2 & 2 & 1\\1 & -1 & 2\\1 & 1 & 1\end{matrix}\right]}$$

解答： $$A=\begin{bmatrix}0 & 2 & 2 \\2 & 0 & 2 \\ 2 & 2 & 0\end{bmatrix} \Rightarrow \det(A-\lambda I)=-(\lambda+2)^2(\lambda-4)=0 \Rightarrow \lambda=-2,4\\ \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 2 & 2 & 2 \\2 & 2 & 2 \\2 & 2 & 2 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \\x_3  \end{bmatrix}=0 \\\qquad \Rightarrow v= x_2 \begin{pmatrix}-1 \\1 \\0 \end{pmatrix} +x_3  \begin{pmatrix}-1 \\0 \\1 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix}-1 \\1 \\0 \end{pmatrix} ,v_2= \begin{pmatrix}-1 \\0 \\1 \end{pmatrix}\\ \lambda_2=4  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -4 & 2 & 2 \\2 & -4 & 2 \\2 & 2 & -4 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \\x_3  \end{bmatrix}=0 \\ \qquad \Rightarrow v= x_3\begin{pmatrix} 1 \\1 \\ 1 \end{pmatrix} , \text{ choose }v_3= \begin{pmatrix}1 \\1 \\ 1 \end{pmatrix}\\ \text{Now, applying Gram-Schmidt process, }e_1={v_1\over |v_1|} =\begin{pmatrix}-1/\sqrt 2 \\1/\sqrt 2 \\0 \end{pmatrix} \\ u_2=v_2-(v_2\cdot e_1)e_1= \begin{pmatrix} -1/2 \\ -1/2 \\ 1 \end{pmatrix} \Rightarrow e_2= {u_2\over |u_2|} =\begin{pmatrix}-1/\sqrt 6 \\-1/\sqrt 6 \\ 2/\sqrt 6 \end{pmatrix} \\ u_3=v_3-(v_3\cdot e_1)e_1-(v_3\cdot e_2)e_2 =\begin{pmatrix}1 \\1 \\ 1 \end{pmatrix} \Rightarrow e_3={u_3\over |u_3|} =\begin{pmatrix}1/\sqrt 3 \\1/\sqrt 3 \\ 1/\sqrt 3 \end{pmatrix} \\ \Rightarrow P=[ e_1\mid e_2\mid e_3]\Rightarrow \bbox[red, 2pt]{P =\begin{bmatrix}-{\sqrt 2\over 2} & -{\sqrt 6\over 6} & {\sqrt 3\over 3}\\{\sqrt 2\over 2} & -{\sqrt 6\over 6} & {\sqrt 3\over 3} \\0& {\sqrt 6\over 3} & {\sqrt 3\over 3} \end{bmatrix}}$$

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