I'm trying to get something like the second Figure; however, the best thing I got is like Figure 1. How can I fix it?

Here is the code:

\{e\}^{n+1} - \{e\}^n = \Delta t \{\dot{e}\}^{n}
+ \beta \Delta t^2 \{\ddot{e}\}^{n+1} \\
+ \Delta t^2 \left( \dfrac{1}{2}-\beta \right) \{\ddot{e}\}^{n}
&\dfrac{1}{\Delta t} \left( \{\dot{e}\}^{n+1}  -  \{\dot{e}\}^{n} \right)=\dfrac{1}{2} \left( \{\ddot{e}\}^{n+1}  +  \{\ddot{e}\}^{n}  \right) 


Fig. 1 (Top) and Fig. 2 (Bottom)

  • Welcome to TeX.SX! I can see no reason for the second type of alignment: there's no relationship between the plus and the minus. – egreg Apr 10 '18 at 8:57
  • Thanks! This is just a simple example of the problem I've faced. I didn't bring the original equation, as it is pretty long. – Mat123 Apr 10 '18 at 9:00

Embed a \tabbedShortstack inside a \Shortstack.

\{e\}^{n+1} - \{e\}^n = \Delta t \{\dot{e}\}^{n}
+ \beta \Delta t^2 \{\ddot{e}\}^{n+1} \\
&+ \Delta t^2 \left( \dfrac{1}{2}-\beta \right) \{\ddot{e}\}^{n}\\
\dfrac{1}{\Delta t} \bigl( \{\dot{e}\}^{n+1}  &-  \{\dot{e}\}^{n} \bigr)=
 \dfrac{1}{2} \left( \{\ddot{e}\}^{n+1}  +  \{\ddot{e}\}^{n}  \right) }}

enter image description here


I would use an align environment, with a \omit\rlap{...} for the first line.

To have the equation number only at the end, I've used \nonumber for the first two rows.

  \omit\rlap{$\{e\}^{n+1} - \{e\}^n = \Delta t \{\dot{e}\}^{n} + \beta \Delta t^2 \{\ddot{e}\}^{n+1}$}\nonumber\\
   &{}+\Delta t^2 \left( \dfrac{1}{2}-\beta \right) \{\ddot{e}\}^{n} \nonumber\\
  \dfrac{1}{\Delta t} \left( \{\dot{e}\}^{n+1}\right. & {}-  \left.\{\dot{e}\}^{n} \right)=\dfrac{1}{2} \left( \{\ddot{e}\}^{n+1}  +  \{\ddot{e}\}^{n}  \right)

enter image description here

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