tex源代码如下:

1 \documentclass[a4paper, 12pt]{article} % Font size (can be 10pt, 11pt or 12pt) and paper size (remove a4paper for US letter paper)  2 \usepackage{amsmath,amsfonts,bm}  3 \usepackage{hyperref}  4 \usepackage{amsthm}   5 \usepackage{amssymb}  6 \usepackage{framed,mdframed}  7 \usepackage{graphicx,color}   8 \usepackage{mathrsfs,xcolor}   9 \usepackage[all]{xy} 10 \usepackage{fancybox}  11 \usepackage{xeCJK} 12 \newtheorem{adtheorem}{定理} 13 \setCJKmainfont[BoldFont=FZYaoTi,ItalicFont=FZYaoTi]{FZYaoTi} 14 \definecolor{shadecolor}{rgb}{
    1.0,0.9,0.9} %背景色为浅红色 15 \newenvironment{theorem} 16 {\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{adtheorem}} 17     {\end{adtheorem}\end{mdframed}\bigskip} 18 \newtheorem*{bdtheorem}{定义} 19 \newenvironment{definition} 20 {\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{bdtheorem}} 21     {\end{bdtheorem}\end{mdframed}\bigskip} 22 \newtheorem*{cdtheorem}{习题} 23 \newenvironment{exercise} 24 {\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{cdtheorem}} 25     {\end{cdtheorem}\end{mdframed}\bigskip} 26 \newtheorem{ddtheorem}{注} 27 \newenvironment{remark} 28 {\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{ddtheorem}} 29     {\end{ddtheorem}\end{mdframed}\bigskip} 30 \newtheorem{edtheorem}{引理} 31 \newenvironment{lemma} 32 {\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{edtheorem}} 33     {\end{edtheorem}\end{mdframed}\bigskip} 34 \usepackage[protrusion=true,expansion=true]{microtype} % Better typography 35 \usepackage{wrapfig} % Allows in-line images 36 \usepackage{mathpazo} % Use the Palatino font 37 \usepackage[T1]{fontenc} % Required for accented characters 38 \linespread{
    1.05} % Change line spacing here, Palatino benefits from a slight increase by default 39  40 \makeatletter 41 \renewcommand\@biblabel[1]{\textbf{#1.}} % Change the square brackets for each bibliography item from '[1]' to '1.' 42 \renewcommand{\@listI}{\itemsep=0pt} % Reduce the space between items in the itemize and enumerate environments and the bibliography 43  44 \renewcommand{\maketitle}{ % Customize the title - do not edit title 45   % and author name here, see the TITLE block 46   % below 47   \renewcommand\refname{参考文献} 48   \newcommand{\D}{\displaystyle}\newcommand{\ri}{\Rightarrow} 49   \newcommand{\ds}{\displaystyle} \renewcommand{\ni}{\noindent} 50   \newcommand{\pa}{\partial} \newcommand{\Om}{\Omega} 51   \newcommand{\om}{\omega} \newcommand{\sik}{\sum_{i=1}^k} 52   \newcommand{\vov}{\Vert\omega\Vert} \newcommand{\Umy}{U_{\mu_i,y^i}} 53   \newcommand{\lamns}{\lambda_n^{^{\scriptstyle\sigma}}} 54   \newcommand{\chiomn}{\chi_{_{\Omega_n}}} 55   \newcommand{\ullim}{\underline{\lim}} \newcommand{\bsy}{\boldsymbol} 56   \newcommand{\mvb}{\mathversion{bold}} \newcommand{\la}{\lambda} 57   \newcommand{\La}{\Lambda} \newcommand{\va}{\varepsilon} 58   \newcommand{\be}{\beta} \newcommand{\al}{\alpha} 59   \newcommand{\dis}{\displaystyle} \newcommand{\R}{
     {\mathbb R}} 60   \newcommand{\N}{
     {\mathbb N}} \newcommand{\cF}{
     {\mathcal F}} 61   \newcommand{\gB}{
     {\mathfrak B}} \newcommand{\eps}{\epsilon} 62   \begin{flushright} % Right align 63     {\LARGE\@title} % Increase the font size of the title 64      65     \vspace{50pt} % Some vertical space between the title and author name 66      67     {\large\@author} % Author name 68     \\\@date % Date 69      70     \vspace{40pt} % Some vertical space between the author block and abstract 71   \end{flushright} 72 } 73  74 % ---------------------------------------------------------------------------------------- 75 %    TITLE 76 % ---------------------------------------------------------------------------------------- 77  78 \title{\textbf{《常微分方程教程》习题2-2,4\\[2em]一个跟踪问题}}  79  80 \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution 81 \renewcommand{\today}{\number\year. \number\month. \number\day} 82 \date{\today} % Date 83  84 % ---------------------------------------------------------------------------------------- 85  86 \begin{document} 87 \maketitle % Print the title section 88  89 % ---------------------------------------------------------------------------------------- 90 %    ABSTRACT AND KEYWORDS 91 % ---------------------------------------------------------------------------------------- 92  93 % \renewcommand{\abstractname}{摘要} % Uncomment to change the name of the abstract to something else 94  95 % \begin{abstract} 96  97 % \end{abstract} 98  99 % \hspace*{
    3,6mm}\textit{关键词:}  % Keywords100 101 % \vspace{30pt} % Some vertical space between the abstract and first section102 103 % ----------------------------------------------------------------------------------------104 %    ESSAY BODY105 % ----------------------------------------------------------------------------------------106 \begin{exercise}[2-2,4]107 跟踪:设某 $A$ 从 $Oxy$  平面的原点出发,沿 $x$ 轴正方向前进;同时某 $B$108 从点 $(0,b)$ 开始跟踪 $A$,即 $B$ 的运动方向永远指向 $A$ 并与 $A$ 保持109 等距 $b$.试求 $B$ 的光滑运动轨迹.110 \end{exercise}111 \begin{proof}[解]112 设在时刻 $t$ 的时候 $A$ 位于 $(f(t),0)$.其中 $f(0)=0$,且 $f(t)$ 是关于113 $t$ 的严格单调增函数.设在时刻 $t$ 的114 $B$ 位于 $(P(t),Q(t))$,其中 $P(0)=0,Q(0)=b$.不妨设 $b\neq 0$,否则 $B$115 的运动将与 $A$ 重合,这是没什么意思的,再根据对称性不妨设 $b>0$.且由于 $B$ 的路径光滑,因此关于116 $t$ 的函数 $P,Q$ 都是连续可微的.由于 $B$ 的方向一直指向 $A$,因此117 \begin{equation}118   \label{eq:10.51}119   (P'(t),Q'(t))=k(f(t)-P(t),-Q(t)).120 \end{equation}121 其中 $k>0$.由于 $A,B$ 间距始终为 $b$,因此122 \begin{equation}123   \label{eq:10.52}124   [P(t)-f(t)]^2+Q(t)^2=b^2.125 \end{equation}126 当 $Q(t)\neq 0$ 时,$Q'(t)$ 也不为0.此时 将(1) 代入 (2) 可得127 \begin{equation}128   \label{eq:11.02}129   (P'(t))^2+(Q'(t))^2=b^2k^2=b^2\frac{Q'(t)^{2}}{Q(t)^{2}}.130 \end{equation}131 于是我们就得到了微分方程132 \begin{equation}133   \label{eq:11.54}134   (\frac{P'(t)}{Q'(t)})^2+1=\frac{b^2}{Q(t)^2}.135 \end{equation}136 也就是137 $$138 (\frac{dP(t)}{dQ(t)})^2+1=\frac{b^2}{Q(t)^2}.139 $$140 也即141 $$142 \frac{dx}{dy}=-\sqrt{(\frac{b}{y})^2-1}.143 $$144 令 $\frac{b}{y}=\cosh a$.其中 $a\in \mathbf{R}^{+}$,于是,145 $$146 \frac{dy}{da}=\frac{-b\tanh a}{\cosh a}.147 $$148 且149 $$150 \frac{dx}{dy}=-\sinh a.151 $$152 因此,153 $$154 \frac{dx}{da}=b(\tanh a)^2=b-b\tanh'a.155 $$156 因此,157 $$158 x=ba-b\tanh a+C.159 $$160 因此,161 $$162 x=b\cosh^{-1}\frac{b}{y}-b\tanh(\cosh^{-1}\frac{b}{y})+C.163 $$164 将初始条件 $x=0,y=b$ 代入,解得 $C=0$.于是 $B$ 的光滑轨迹为165 $$166 x=b\cosh^{-1}\frac{b}{y}-b\tanh(\cosh^{-1}\frac{b}{y}).167 $$168 通过这个方程,我们发现 $B$ 的运动轨迹和 $A$ 的运动无关!\\169 170 当 $Q(t)=0$ 时,易得 $B$ 已经和 $A$ 同在 $x$ 轴上运动.171 \end{proof}172 % ----------------------------------------------------------------------------------------173 %    BIBLIOGRAPHY174 % ----------------------------------------------------------------------------------------175 176 \bibliographystyle{unsrt}177 178 \bibliography{sample}179 180 % ----------------------------------------------------------------------------------------181 \end{document}