I have a system of linear non-autonomous, linear, homogeneous equations: $$ \begin{bmatrix} \dot{x}(t) \\ \dot{y}(t) \end{bmatrix} = \begin{bmatrix} a & e^{-i b t} \\ e^{i b t} & 0 \end{bmatrix} \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} $$ with initial conditions $x(0) = x_{0}$ and $y(0) = y_{0}$. I am able to easily get an analytic solution for this system via Wolfram Mathematica, however I would like to be able to reproduce this result with pen and paper but I am at a loss for the approach.
My first thought was to try solving using the matrix exponential: $$\mathbf{X}(t) = \exp\left(\int_{0}^{t}\mathbf{A}(\tau) \ {\rm{d}}\tau\right)\mathbf{X}(0) \text{,}$$ however this does not work because (as I later realised) this approach works only if $\mathbf{A}(t)$ commutes for different times. I have also flirted with the time-ordered exponential approach (borrowing from Quantum Mechanics), but this does not replicate my Mathematica result (which I trust).
Write $X(t) = \exp(ibt)x(t)$. Then
$$ \dot{X}(t) = ib X(t) + \exp(ibt)\dot{x}(t) = ib X(t) + a X(t) + y(t) $$
and
$$ \dot{y}(t) = X(t) $$
So the system now solves
$$ \begin{bmatrix} \dot{X}(t) \\ \dot{y}(t) \end{bmatrix} = \begin{bmatrix} a + ib & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} X(t) \\ y(t) \end{bmatrix}$$
which is an autonomous system.