This problem is related to a pendulum problem. Let $X := \left(\begin{matrix} x(t)\\y(t)\end{matrix}\right)$ and, $T,m,L>0$ real numbers, define also the matrix
$$A := \left(\begin{matrix} 0&-T/mL\\-T/mL &0\end{matrix}\right) = \frac{-T}{mL}\sigma_1 = -\omega_0\sigma_1$$
Where we defined $\omega_0: T/mL$. We have that $\sigma_1:= \left(\begin{matrix} 0&1\\1 &0\end{matrix}\right)$ is one of the Pauli matrices. So we get that $\sigma_1^2 = I$ the identity matrix $2\times 2$.
My problem is: solve the above differential equation, for $X(0) := X_0$, $$\ddot{X} = AX$$
What I tried was to use that, as $A$ is a constant matrix we get that $X(t) = X_0e^{At}$. Then the problem is to find $e^{At}$.
$$e^{At} = \sum_{n=0}^\infty\frac{(At)^n}{n!} = \sum_{n=0}^\infty\frac{(-\omega_0t)^n\sigma_1^n}{n!} = I\cdot \sum_{n=0}^\infty\frac{(\omega_0t)^{2n}}{(2n)!}-\sigma_1\sum_{n=0}^\infty\frac{(\omega_0t)^{2n+1}}{(2n+1)!} \implies$$
$$e^{At} = I\cosh(\omega_0t)-\sigma_1\sinh(\omega_0t) = \left(\begin{matrix} \cosh(\omega_0t)&-\sinh(\omega_0t)\\-\sinh(\omega_0t) &\cosh(\omega_0t)\end{matrix}\right)$$
But this is clearly wrong. Set $X_0 =\left(\begin{matrix} 1\\0\end{matrix} \right)$ and we get for $x(t)$ and $y(t)$
$$x(t) = \cosh(\omega_0t)$$
$$y(t) = -\sinh(\omega_0t)$$
But $\ddot{x}(t) = \omega_0^2\cosh(\omega_0 t)\neq -\omega_0y(t)$
Note, that you cannot use the matrix exponential for a second order system. The matrix exponential is only applicable for first order linear time-invariant systems (first order LTI). Your ODE is time-invariant and linear but is not of first order.
If you had a fist order linear time-variant (first order LTV) system you could apply the Peano-Baker-series.
Another way to solve your problem would be to put it into the first-order form by introducing intermediate variables such that the resulting system is a linear first order system which is time-invariant and then apply the matrix exponential method.