I am considering the following equation


$$\begin{pmatrix} -\frac{d}{dx} + \lambda \sin(2\pi x) & \lambda - \lambda \cos(2\pi x) \\ -\lambda-\lambda \cos(2\pi x) & -\frac{d}{dx} - \lambda \sin(2\pi x) \end{pmatrix}\varphi(x)=0$$


and I am wondering if there is an explicit characterization of $\lambda \neq 0$ for which there exist $1$-periodic solutions to this ODE.


This ODE, though matrix-valued has a nice symmetry in the sense that the off-diagonal entries have only even Taylor polynomials whereas the diagonal ones have odd ones.


However, it is of course difficult to say based on a Taylor series, if this will yield a periodic function.


*Observation:*


Close to zero, i.e. $x=0$, the solution should heuristically look like $\varphi(x)=(0,e^{-\lambda \pi x^2}),$ as it solves the equation "to leading-order"



Matrix-valued ordinary differential equation with symmetry
Your equation in fact admits a simple explicit solution. As noted in the previous answer, the general solution is encoded in a matrix ODE,
$$
M'(x) = \lambda A(x)M(x), \qquad M(0)=I\,.
$$
Now, the simplification becomes apparent if change to a new basis, using the $x$-dependent rotation
$$
R(x) = \begin{pmatrix} \cos(\pi x) & -\sin(\pi x) \\ \sin(\pi x) & \cos(\pi x)\end{pmatrix}.
$$
Define $N(x)$ by $M(x)=R(x)N(x)$, and the ODE becomes
$$
N' = \left(\lambda R^{-1}AR-R^{-1}R' \right)N = \begin{pmatrix} 0 & \pi \\ -\pi-2\lambda & 0\end{pmatrix} N.
$$


Writing $\lambda = \frac{\pi}{2}(\mu^2-1)$, the solution is
$$
N(x) = \begin{pmatrix}
 \cos (\pi\mu x) & \frac{1}{\mu} \sin (\pi\mu x) \\
 -\mu \sin (\pi\mu x) & \cos (\pi\mu x)
\end{pmatrix} \implies M(1) = \begin{pmatrix}
 -\cos (\pi\mu) & -\frac{1}{\mu} \sin(\pi\mu) \\
 \mu \sin(\pi\mu) & -\cos(\pi\mu)
\end{pmatrix}.
$$
We get periodic solutions when $\mu=2n+1$, $n=1,2,3,\ldots$, so $\lambda=2\pi n(n+1)$.


