tex文档:

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,epigraph}   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 \usepackage{CJKutf8} 13 \newtheorem*{adtheorem}{定理} 14 % \setCJKmainfont[BoldFont=FZYaoTi,ItalicFont=FZYaoTi]{FZYaoTi} 15 \definecolor{shadecolor}{rgb}{
    1.0,0.9,0.9} %背景色为浅红色 16 \newenvironment{theorem} 17 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{adtheorem}} 18     {\end{adtheorem}\end{mdframed}\bigskip} 19 \newtheorem*{bdtheorem}{定义} 20 \newenvironment{definition} 21 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{bdtheorem}} 22     {\end{bdtheorem}\end{mdframed}\bigskip} 23 \newtheorem*{cdtheorem}{习题} 24 \newenvironment{exercise} 25 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{cdtheorem}} 26     {\end{cdtheorem}\end{mdframed}\bigskip} 27 \newtheorem*{ddtheorem}{注} 28 \newenvironment{remark} 29 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{ddtheorem}} 30     {\end{ddtheorem}\end{mdframed}\bigskip} 31 \newtheorem*{edtheorem}{引理} 32 \newenvironment{lemma} 33 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{edtheorem}} 34     {\end{edtheorem}\end{mdframed}\bigskip} 35 \newtheorem*{pdtheorem}{例} 36 \newenvironment{example} 37 {\bigskip\begin{mdframed}[backgroundcolor=gray!40,rightline=false,leftline=false,topline=false,bottomline=false]\begin{pdtheorem}} 38     {\end{pdtheorem}\end{mdframed}\bigskip} 39  40 \usepackage[protrusion=true,expansion=true]{microtype} % Better typography 41 \usepackage{wrapfig} % Allows in-line images 42 \usepackage{mathpazo} % Use the Palatino font 43 \usepackage[T1]{fontenc} % Required for accented characters 44 \linespread{
    1.05} % Change line spacing here, Palatino benefits from a slight increase by default 45  46 \makeatletter 47 \renewcommand\@biblabel[1]{\textbf{#1.}} % Change the square brackets for each bibliography item from '[1]' to '1.' 48 \renewcommand{\@listI}{\itemsep=0pt} % Reduce the space between items in the itemize and enumerate environments and the bibliography 49  50 \renewcommand{\maketitle}{ % Customize the title - do not edit title 51   % and author name here, see the TITLE block 52   % below 53   \renewcommand\refname{参考文献} 54   \newcommand{\D}{\displaystyle}\newcommand{\ri}{\Rightarrow} 55   \newcommand{\ds}{\displaystyle} \renewcommand{\ni}{\noindent} 56   \newcommand{\pa}{\partial} \newcommand{\Om}{\Omega} 57   \newcommand{\om}{\omega} \newcommand{\sik}{\sum_{i=1}^k} 58   \newcommand{\vov}{\Vert\omega\Vert} \newcommand{\Umy}{U_{\mu_i,y^i}} 59   \newcommand{\lamns}{\lambda_n^{^{\scriptstyle\sigma}}} 60   \newcommand{\chiomn}{\chi_{_{\Omega_n}}} 61   \newcommand{\ullim}{\underline{\lim}} \newcommand{\bsy}{\boldsymbol} 62   \newcommand{\mvb}{\mathversion{bold}} \newcommand{\la}{\lambda} 63   \newcommand{\La}{\Lambda} \newcommand{\va}{\varepsilon} 64   \newcommand{\be}{\beta} \newcommand{\al}{\alpha} 65   \newcommand{\dis}{\displaystyle} \newcommand{\R}{
     {\mathbb R}} 66   \newcommand{\N}{
     {\mathbb N}} \newcommand{\cF}{
     {\mathcal F}} 67   \newcommand{\gB}{
     {\mathfrak B}} \newcommand{\eps}{\epsilon} 68   \begin{flushright} % Right align 69     {\LARGE\@title} % Increase the font size of the title 70      71     \vspace{50pt} % Some vertical space between the title and author name 72      73     {\large\@author} % Author name 74     \\\@date % Date 75      76     \vspace{40pt} % Some vertical space between the author block and abstract 77   \end{flushright} 78 } 79  80 % ---------------------------------------------------------------------------------------- 81 %    TITLE 82 % ---------------------------------------------------------------------------------------- 83 \begin{document} 84 \begin{CJK}{UTF8}{gkai} 85   \title{\textbf{《常微分方程教程》\cite{dinglichang}习题2.4.1,(4)}} 86   % \setlength\epigraphwidth{
    0.7\linewidth} 87   \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学 88         号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution 89   \renewcommand{\today}{\number\year. \number\month. \number\day} 90   \date{\today} % Date 91    92   % ---------------------------------------------------------------------------------------- 93    94    95   \maketitle % Print the title section 96    97   % ---------------------------------------------------------------------------------------- 98   %    ABSTRACT AND KEYWORDS 99   % ----------------------------------------------------------------------------------------100   101   % \renewcommand{\abstractname}{摘要} % Uncomment to change the name of the abstract to something else102   103   % \begin{
    abstract}104   105   % \end{
    abstract}106   107   % \hspace*{
    3,6mm}\textit{关键词:} % Keywords108   109   % \vspace{30pt} % Some vertical space between the abstract and first section110   111   % ----------------------------------------------------------------------------------------112   %    ESSAY BODY113   % ----------------------------------------------------------------------------------------114   \begin{exercise}[2.4.1,(4)]115     求解下列微分方程:116 $$117 y'=x^3y^3-xy.118 $$119 \end{exercise}120 \begin{proof}[解]121 即为122 $$123 \frac{dy}{dx}=x^3y^3-xy.124 $$125 这是个 Bernoulli 方程.当 $y\neq 0$ 时,两边同时除以 $y^3$,可得126 $$127 \frac{
    1}{y^3}\frac{dy}{dx}+\frac{
    1}{y^2}x-x^3=0.128 $$129 令 $z=y^{-2}$,则130 $$131 \frac{dz}{dx}=-2y^{-3}\frac{dy}{dx},132 $$133 因此134 $$135 \frac{dz}{dx}-2zx+2x^3=0.136 $$137 这是个关于 $z,x$ 的一阶线性方程.可化为138 $$139 dz+(2x^3-2zx)dx=0.140 $$141 乘以积分因子 $u(x)$,则142 $$143 udz+u(2x^3-2zx)dx=0.144 $$145 令 146 $$147 \frac{du}{dx}=-2xu,148 $$149 不妨令 $u=e^{\int -2xdx}$.因此我们得到恰当方程150 $$151 e^{\int -2xdx}dz+e^{\int -2xdx}(2x^3-2zx)dx=0.152 $$153 其中两个 $e^{\int -2xdx}$ 是同一个函数.设存在二元函数 $\phi(x,y)$ 使得154 $$155 \frac{\pa\phi}{\pa z}=e^{\int -2xdx}\ri \phi=ze^{\int -2xdx}+f(x).156 $$157 因此158 $$159 -2xze^{\int -2xdx}+f'(x)=e^{\int -2xdx}(2x^3-2zx).160 $$161 可得162 $$163 f'(x)=2x^3e^{\int -2xdx}\ri f(x)=-x^2e^{\int -2xdx}-e^{\int -2xdx}+C.164 $$165 因此可得通积分为166 $$167 \phi\equiv ze^{\int -2xdx}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0.168 $$169 其中三个 $e^{\int -2xdx}$ 都是同一个函数.将 $z=y^{-2}$ 代入,可得170 $$171 \frac{e^{\int -2xdx}}{y^2}-x^2e^{\int -2xdx}-e^{\int -2xdx}+C=0.172 $$173 其中三个 $e^{\int -2xdx}$ 都是同一个函数.不妨设 $\int -2xdx=-x^2+D$,因174 此可得175 $$176 \frac{e^{-x^2}}{y^2}-x^2e^{-x^2}-e^{-x^2}+C'=0.177 $$178 而当 $y=0$ 时,可得曲线为 $y=0$.179 \end{proof}180 % ----------------------------------------------------------------------------------------181 %    BIBLIOGRAPHY182 % ----------------------------------------------------------------------------------------183   184 \bibliographystyle{unsrt}185   186 \bibliography{sample}187   188 % ----------------------------------------------------------------------------------------189 \end{CJK}190 \end{document}