如何以良好的形式编写这些系统

Question

我得到了这个三元方程组。我想把它们写成漂亮的形式

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{align}
 e^{y}[m_0 \sin(c)-m_1\cos(kh)]=&0 \\
        e^{-\alpha h}[m_0 \sin(kh)+m_2\tan(kh)]&=0 \\
           m_0\left[2k\kappa  \sin(kh)\cosh(yu)-2\alpha\kappa \cos (k h)\sinh (\alpha h)\right]
\nonumber \\+m_1[-k+\alpha \kappa e^{yu} \cos(c)+k e^{y} \cos(kh)]\nonumber \\+m_2[w+\alpha  e^{-y}
\tan(kh)-k  e^{-y}\sin(c)]&=0
\end{align}

\end{document}

Answer 1

我会避免对齐那些=0会让不太重要的部分过于突出的部分。以下是两个建议：

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\begin{align}
& e^{y}[m_0 \sin(c)-m_1\cos(kh)]=0 \\[1ex]
& e^{-\alpha h}[m_0 \sin(kh)+m_2\tan(kh)]=0 \\[1ex]
& \!\begin{aligned}[b]
  & m_0[2k\kappa \sin(kh)\cosh(yu)-2\alpha\kappa \cos(kh)\sinh(\alpha h)] \\
  & \qquad+m_1[-k+\alpha \kappa e^{yu} \cos(c)+k e^{y} \cos(kh)] \\
  & \qquad+m_2[w+\alpha e^{-y}\tan(kh)-k e^{-y}\sin(c)]=0
  \end{aligned}
\end{align}

\begin{align}
& e^{y}[m_0 \sin(c)-m_1\cos(kh)]=0 \\[1ex]
& e^{-\alpha h}[m_0 \sin(kh)+m_2\tan(kh)]=0 \\[1ex]
& \!\begin{aligned}[b]
  m_0[2k\kappa \sin(kh)\cosh(yu)-2\alpha\kappa \cos(kh)\sinh(\alpha h)] \\
  {}+m_1[-k+\alpha \kappa e^{yu} \cos(c)+k e^{y} \cos(kh)] \\
  {}+m_2[w+\alpha e^{-y}\tan(kh)-k e^{-y}\sin(c)]&=0
  \end{aligned}
\end{align}

\end{document}

Answer 2

也许是这个？

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{align}
 e^{y}[m_0 \sin(c)-m_1\cos(kh)]&=0 \\
        e^{-\alpha h}[m_0 \sin(kh)+m_2\tan(kh)]&=0 \\[1ex]
          \begin{aligned}[b] m_0 [2k\kappa \sin(kh)\cosh(yu) -2\alpha\kappa \cos (k h)\sinh (\alpha h)]&
 \\{}+m_1 [-k +\alpha \kappa e^{yu} \cos(c) +k e^{y} \cos(kh)]& \\{}+m_2 [w +\alpha e^{-y} \tan(kh) -k e^{-y}\sin(c)]& \end{aligned}&=0
\end{align}

\end{document} 

Answer 3

使用alignat。这是我的尝试

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\begin{alignat}{4}
 e^{y}         & [m_0 \sin(c)                     & - & m_1\cos(kh)] &=0 \\
 e^{-\alpha h} & [m_0 \sin(kh)                    & + & m_2\tan(kh)] &=0 \\
           m_0 & [2k\kappa \sin(kh)\cosh(yu)      & - & 2\alpha\kappa \cos (k h)\sinh (\alpha h)] \nonumber \\
          +m_1 & [-k+\alpha \kappa e^{yu} \cos(c) & + & k e^{y} \cos(kh)] \nonumber \\
          +m_2 & [w+\alpha  e^{-y} \tan(kh)       & - & k  e^{-y}\sin(c)] &=0
\end{alignat}

\end{document}

另外，检查这个问题回答有关清理方程式的更多技巧。