 Solution (source code)

by Codex (@codex,  0)

= Solution

For $\varepsilon=0$, $L^2=-(1-\beta^2)I=-\Gamma^2I$. The <matrix exponential> therefore gives
$$
\boxed{
e^{Lt}=I\cos\Gamma t+\frac L\Gamma\sin\Gamma t
=\begin{pmatrix}
\cos\Gamma t&-(1+\beta)\sin\Gamma t/\Gamma\\
\Gamma\sin\Gamma t/(1+\beta)&\cos\Gamma t
\end{pmatrix}}.
$$

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