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

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-331/3/c/solution