= Solution
For a constant matrix, the <matrix exponential> series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a <operator norm>. Termwise <derivatives> therefore give
$$
\frac{d}{dt}e^{tM}=\sum_{n=1}^{\infty}\frac{t^{n-1}M^n}{(n-1)!}=Me^{tM},\qquad e^{0M}=I.
$$
Hence $\mathbf x(t)=e^{tM}\mathbf x_0$ solves the <initial value problem>. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the <matrix exponential> into its even and odd powers proves
$$
\boxed{e^{tM}=I\cos t+M\sin t=\begin{pmatrix}\cos t+2\sin t&5\sin t\\-\sin t&\cos t-2\sin t\end{pmatrix}.}
$$
This is the <matrix exponential when the square is minus the identity>. If $\mathbf x_0=(x_0,y_0)^{\mathsf T}$, the <general solution> is
$$
\boxed{x(t)=x_0\cos t+(2x_0+5y_0)\sin t,\qquad y(t)=y_0\cos t-(x_0+2y_0)\sin t.}
$$
The constants in part (a) are $\alpha=x_0/5$ and $\beta=y_0+2x_0/5$, which makes the two forms identical.
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