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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 else 102 103 % \begin{abstract} 104 105 % \end{abstract} 106 107 % \hspace*{3,6mm}\textit{关键词:} % Keywords 108 109 % \vspace{30pt} % Some vertical space between the abstract and first section 110 111 % ---------------------------------------------------------------------------------------- 112 % ESSAY BODY 113 % ---------------------------------------------------------------------------------------- 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 % BIBLIOGRAPHY 182 % ---------------------------------------------------------------------------------------- 183 184 \bibliographystyle{unsrt} 185 186 \bibliography{sample} 187 188 % ---------------------------------------------------------------------------------------- 189 \end{CJK} 190 \end{document}