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Display information for equation id:math.1968.5 on revision:1968
* Page found: Der Satz von Liouville (eq math.1968.5)
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Hash: 4aa536345e39106ce83f864fd904b1b4
TeX (original user input):
\begin{align}
& \bar{x}(t)=\sum\limits_{n}{{}}\frac{{{\left[ \left( t-{{t}_{0}} \right)A \right]}^{n}}}{n!}{{{\bar{x}}}_{0}}=\left[ 1\cos {{\omega }_{0}}(t-{{t}_{0}})+\frac{A}{{{\omega }_{0}}}\sin {{\omega }_{0}}(t-{{t}_{0}}) \right]{{{\bar{x}}}_{0}} \\
& =\left( \begin{matrix}
\cos {{\omega }_{0}}(t-{{t}_{0}}) & \frac{1}{{{\omega }_{0}}}\sin {{\omega }_{0}}(t-{{t}_{0}}) \\
-{{\omega }_{0}}\sin {{\omega }_{0}}(t-{{t}_{0}}) & \cos {{\omega }_{0}}(t-{{t}_{0}}) \\
\end{matrix} \right){{{\bar{x}}}_{0}} \\
\end{align}
TeX (checked):
{\begin{aligned}&{\bar {x}}(t)=\sum \limits _{n}{}{\frac {{\left[\left(t-{{t}_{0}}\right)A\right]}^{n}}{n!}}{{\bar {x}}_{0}}=\left[1\cos {{\omega }_{0}}(t-{{t}_{0}})+{\frac {A}{{\omega }_{0}}}\sin {{\omega }_{0}}(t-{{t}_{0}})\right]{{\bar {x}}_{0}}\\&=\left({\begin{matrix}\cos {{\omega }_{0}}(t-{{t}_{0}})&{\frac {1}{{\omega }_{0}}}\sin {{\omega }_{0}}(t-{{t}_{0}})\\-{{\omega }_{0}}\sin {{\omega }_{0}}(t-{{t}_{0}})&\cos {{\omega }_{0}}(t-{{t}_{0}})\\\end{matrix}}\right){{\bar {x}}_{0}}\\\end{aligned}}
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