multicol-demo.tex

\documentclass[9pt,oneside,openany]{ctexbook}
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\usepackage{hlist}
%\usepackage{lipsum}
\usepackage{xcolor}
\usepackage{balance}
\usepackage{multicol}
\usepackage{amssymb}%打小黑格
\usepackage{enumerate}
\usepackage{enumitem}%设置item环境
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\usepackage[utf8]{inputenc}
\title{\textbf{\LARGE{数学专业英语期末翻译作业}}} %大括号里填写标题,textbf加粗
\author{\textbf{学院：\underline{应用数学学院}}\\ %\\表示换行
 \textbf{班级：\underline{\  \ \, }}\\    %\表示空格,下quad同
 \textbf{姓名：\underline{\quad \ \quad \  }}\\
 \textbf{学号：\underline{\ \,}}\\ \\
 \textbf{翻译页码：P237-238}
}%end author
\date{\today} % 自动生成日期
\usepackage{booktabs}%线条粗细
\begin{document}
 \par\addtolength{\leftskip}{100pt}
 {\small
 \noindent This formula is more complicated than it needs to be. If we combine $C=C_{1}-2C_{2}$, and $5C_{3}$ into a single arbitrary constant $C=C_{1}-2C_{2}+5C_{3}$, the formulas simplifies to

 \noindent 这个公式比实际情况还要复杂。如果我们把$C=C_{1}-2C_{2}$,和 $5C_{3}$合并到一个任意常数的式子$C=C_{1}-2C_{2}+5C_{3}$，这个公式可以被简化成
 $$ y=\frac{x^3}{3}-x^2+5x+C$$
 and $still$ gives all the possible antiderivatives there are. For this reason, we recommend that you go right to the final form even if you elect to integrate term-by-term. Write

 \noindent 并且依然给所有可能的不定积分。基于这个原因，我们推荐你直接看最后的形式，即使你需要逐项的整合。写
 $$\int(x^2-2x+5)dx=\int x^2dx-\int 2xdx+\int 5dx$$
 $$=\frac{x^3}{3}-x^2+5x+C.$$
 Find the simpliest antiderivative you can for each part and add the arbitary constant of integration at the end.

 \noindent 为每个部分找出它最简形式的反导数，然后在最后取一个任意的正常数。}
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 \par
 \addtolength{\leftskip}{-100pt}
 \noindent\hspace*{-0.3em}\rule[-1pt]{17.85cm}{0.04em}
 \hspace*{-0.2em}\colorbox[rgb]{0,0.28,0.56}{\textcolor{white}{\Large{\textbf{EXERCISES}}}} \
 \textcolor[rgb]{0.50,0.25,0.25}{\Large{\textbf{4.7}}}
 \begin{multicols}{2}
  \noindent{\color[rgb]{0,0.28,0.56}{\footnotesize\textbf{Finding Antiderivatives}}}\\
  {\color[rgb]{0,0.28,0.56}{\footnotesize\textbf{寻找反导数}}}\\
{\scriptsize
 In Exercises 1-16, find an antiderivative for each function. Do as many as you can mentally. Check your answer by differentiation.}\\
{\scriptsize 在练习1-16中，为每个函数寻找一个不定积分。在你的能力范围内尽可能寻找。并且通过微分法验证你的答案。
}
 \begin{enumerate}[leftmargin=5mm]
  \begin{multicols}{3}
  {\footnotesize
       \item [\bf1.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf2.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf3.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf4.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf5.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf6.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf7.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf8.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf9.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf10.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf11.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf12.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf13.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf14.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf15.]\textbf{a.} {\textbf{$y=2x$}}
    \item [\bf16.]\textbf{a.} {\textbf{$y=2x$}}
    %第二栏
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    \item [\bf b.] {\textbf{$y=2x$}}
    %第三栏
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
    \item [\bf c.] {\textbf{$y=2x$}}
   }
  \end{multicols}
 \end{enumerate}
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\noindent{\color[rgb]{0,0.28,0.56}{\footnotesize\textbf{Finding Indefinite Integrals}}}\\
{\color[rgb]{0,0.28,0.56}{\footnotesize\textbf{寻找反导数}}}\\
{\scriptsize
In Exercises 17-56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.\\
在练习17-56中，为每个函数寻找一个不定积分。在你的能力范围内尽可能寻找。并且通过微分法验证你的答案。}
\begin{itemize}[leftmargin=4.8mm]
 \begin{multicols}{2}

   \item [\bf17.]{\textbf{$\int (sin2x-csc^{2}x)dx$}}
   \item [\bf19.]{\textbf{$\int \frac{1+cos4t}{2}dt$}}
   \item [\bf21.]{\textbf{$\int 3x^{\sqrt{3}}dx$}}
   \item [\bf23.]{\textbf{$\int (1+tan^{2}\theta )d\theta$}}
   \item [\bf25.]{\textbf{$\int cot^{2}xdx$}}
   \item [\bf27.]{\textbf{$\int cos\theta (tan\theta+sec \theta)d\theta$}}
   \item [\bf29.]{\textbf{$\int (2cos2x-3sin3x)dx$}}
   \item [\bf31.]{\textbf{$\int \frac{1-cos6t}{2}dt$}}
   \item [\bf33.]{\textbf{$\int x^{\sqrt{2}-1}dx$}}
   \item [\bf35.]{\textbf{$\int (2=tan^{2}\theta)d\theta$}}
   \item [\bf37.]{\textbf{$\int (1-cot^{2}x)dx$}}
   \item [\bf39.]{\textbf{$\int \frac{csc\theta}{csc\theta -sin\theta}$}}
   \item [\bf41.]{\textbf{$\int (sin2x-csc^{2}x)dx$}}
   \item [\bf43.]{\textbf{$\int (4secxtanx-2(secx)^2{2}dx$}}
   \item [\bf18.]{\textbf{$\int 3x^{\sqrt{3}}dx$}}
   \item [\bf20.]{\textbf{$\int (1+tan^{2}\theta )d\theta$}}
   \item [\bf22.]{\textbf{$\int cot^{2}xdx$}}
   \item [\bf24.]{\textbf{$\int cos\theta (tan\theta+sec \theta)d\theta$}}
   \item [\bf26.]{\textbf{$\int (2cos2x-3sin3x)dx$}}
   \item [\bf28.]{\textbf{$\int \frac{1-cos6t}{2}dt$}}
   \item [\bf30.]{\textbf{$\int x^{\sqrt{2}-1}dx$}}
   \item [\bf32.]{\textbf{$\int (2=tan^{2}\theta)d\theta$}}
   \item [\bf34.]{\textbf{$\int (1-cot^{2}x)dx$}}
   \item [\bf36.]{\textbf{$\int \frac{csc\theta}{csc\theta -sin\theta}$}}
   \item [\bf38.]{\textbf{$\int (1-cot^{2}x)dx$}}
   \item [\bf40.]{\textbf{$\int \frac{csc\theta}{csc\theta -sin\theta}$}}
   \item [\bf42.]{\textbf{$\int \frac{csc\theta}{csc\theta -sin\theta}$}}
 \end{multicols}
\end{itemize}
\end{multicols}
% \clearpage
\begin{multicols}{2}
\begin{enumerate}[leftmargin=4.8mm]
\item [\bf44.]\small{\textbf{$\int \frac{1}{2}(csc^{2}x-cscxcotx)dx$}}%上一页自动顺延下来的。
\begin{multicols}{2}
 {\balance
  \item [\bf45.]\small{\textbf{$\int (sin2x-csc^{2}x)dx$}}
  \item [\bf47.]\small{\textbf{$\int \frac{1+cos4t}{2}dt$}}
  \item [\bf49.]\small{\textbf{$\int 3x^{\sqrt{3}}dx$}}
  \item [\bf51.]\small{\textbf{$\int (1+tan^{2}\theta )d\theta$}}
  \item [\bf53.]\small{\textbf{$\int cot^{2}xdx$}}
  \item [\bf55.]\small{\textbf{$\int cos\theta (tan\theta+sec \theta)d\theta$}}
  \item [\bf46.]\small{\textbf{$\int (2cos2x-3sin3x)dx$}}
  \item [\bf48.]\small{\textbf{$\int \frac{1-cos6t}{2}dt$}}
  \item [\bf50.]\small{\textbf{$\int x^{\sqrt{2}-1}dx$}}
  \item [\bf52.]\small{\textbf{$\int (2=tan^{2}\theta)d\theta$}}
  \item [\bf54.]\small{\textbf{$\int (1-cot^{2}x)dx$}}
  \item [\bf56.]\small{\textbf{$\int \frac{csc\theta}{csc\theta -sin\theta}$}}
 }
\end{multicols}
\end{enumerate}
\end{multicols}
\end{document}


\documentclass{article}
\usepackage[overload]{empheq}
\begin{document}

\begin{subequations}
  \begin{empheq}[left = {\empheqlbrace}]{alignat = 2}
    u_t - \Delta u &= 0 & \qquad & \text{in \(U_T\)}, \\
    u              &= 0 & \qquad & \text{on \(\partial U\times [0, T]\)}, \\
    u              &= g & \qquad & \text{on \(U\times \{t = 0\}\)}.
  \end{empheq}
\end{subequations}

\end{document}