我正在使用一些由脚本自动生成的方程式，并使用breqn函数进行排版。
使用dgroup会导致我的方程编号位于左侧而不是右侧。我按此顺序调用包，据我从文档中看到的那样，这应该是正确的。
\usepackage{amsmath}
\usepackage{flexisym}
\usepackage{breqn}
\usepackage{bm}
\usepackage{amsfonts}
正在排版的方程式是
\begin{dgroup}
\begin{dmath*}
0=\frac{\partial }{\partial t} y_{2}\left(t\right)-\frac{\sin\left(y_{3}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{3}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{3}\left(t\right)\right)\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{3}\left(t\right)\right)}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-y_{5}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{3}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{1}\left(t\right)}{\cos\left(y_{3}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{4}\left(t\right)-y_{3}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\sin\left(y_{4}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{2}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{4}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{4}\left(t\right)\right)\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{4}\left(t\right)\right)}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{4}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{1}\left(t\right)}{\cos\left(y_{4}\left(t\right)\right)}
\end{dmath*}
\end{dgroup}
答案1
不是一个解决方案，但这似乎是 中的一个已知错误breqn，请参阅包装手册：
因此，目前您无法按照自己的方式实现目标。我会检查是否可以找到解决方法。
好的，这是一个不太优雅的解决方法：
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{flexisym}
\usepackage{breqn}
\usepackage{bm}
\begin{document}
\begin{equation}
\parbox{.9\linewidth}{%
\begin{dgroup*}
\begin{dmath*}
0=\frac{\partial }{\partial t} y_{2}\left(t\right)-\frac{\sin\left(y_{3}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{3}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{3}\left(t\right)\right)\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{3}\left(t\right)\right)}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-y_{5}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{3}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{1}\left(t\right)}{\cos\left(y_{3}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{4}\left(t\right)-y_{3}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\sin\left(y_{4}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{2}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{4}\left(t\right)\right)\cdot y_{5}\left(t\right)}{6371}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=\frac{\cos\left(y_{4}\left(t\right)\right)\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{4}\left(t\right)\right)}-\frac{\partial }{\partial t} y_{1}\left(t\right)
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{2}\left(t\right)}{\sin\left(y_{4}\left(t\right)\right)}
\end{dmath*}
\begin{dmath*}
0=y_{6}\left(t\right)-\frac{6371\cdot \frac{\partial }{\partial t} y_{1}\left(t\right)}{\cos\left(y_{4}\left(t\right)\right)}
\end{dmath*}
\end{dgroup*}
}
\end{equation}
\end{document}
