Properties of the solutions for linear system with variable coefficients Let us consider the following $2\times 2$ linear system of ODEs with variables coefficients:
$$
\dfrac{d}{dt}\left(\begin{matrix}
y_1(t)\\ y_2(t)
\end{matrix}\right)= \left( \begin{matrix} 
0 & f_{12}(t) \\ f_{21}(t) & 0 
\end{matrix}\right) \left( \begin{matrix} 
y_1(t) \\ y_2(t)
\end{matrix}\right),
$$
where $f_{12}(t),f_{21}(t):\mathbb{R}\to\mathbb{R}$ are some fixed functions. I am wondering if it is possible to prove something like, if $f_{12}$ and $f_{21}$ have some parity properties, lets say $f_{12}$ is even and $f_{21}$ is odd, then the previous system has exactly two linearly independent solutions $$
(y_1^1(t),y_2^1(t)) \quad \hbox{and} \quad (y_1^2(t),y_2^2(t)),
$$
and they enjoy some simmetry properties as well, say $y_1^1(t)$ is even and $y_2^1(t)$ is odd, or things like that. Is this always possible to do, even with other parities, for example $f_{12}$ odd and $f_{21}$ also odd? Or maybe to say that at least one of the fundamental solutions has some simmetry?
 A: Idea to try and answer: (might be a bit long for a comment)
The system of equations seems to be
$$\frac{d y_1}{dt}=f_{12}y_2 \\ \frac{d y_2}{dt}=f_{21}y_1$$
If you differentiate again, you can get second order linear homogeneous differential equations for each function.
$$\frac{d^2 y_1}{dt^2}=\frac{1}{f_{12}}\frac{d f_{12}}{dt}\frac{d y_1}{dt}+f_{12}f_{21}y_1$$
and
$$\frac{d^2 y_2}{dt^2}=\frac{1}{f_{21}}\frac{d f_{21}}{dt}\frac{d y_2}{dt}+f_{21}f_{12}y_2$$
or in the form
$$\frac{d}{dt}\left(\frac{dy_1}{dt}\cdot \frac{1}{f_{12}}\right)=f_{21}y_1$$
$$\frac{d}{dt}\left(\frac{dy_2}{dt}\cdot \frac{1}{f_{21}}\right)=f_{12}y_2$$
Maybe consider $z_j(t)=y_j(-t)=y_j(g(t))$ with $g(t)=-t$ and trying to apply the chain to rule to see if $z_j$ satisfies a similar second order homogenous equation.
That is
$$\frac{d z_j}{dt}=\frac{d z_j}{dg}\frac{dg}{dt}=-\frac{d y_j(g)}{dg}=-f_{jk}y_k(g)=-f_{jk}z_k$$
where $j=1,2$ and $k=3-j$. Then
$$\frac{d^2 z_j}{dt^2}=-\frac{d f_{jk}}{dt}z_k-f_{jk}\frac{d z_k}{dt}=+\frac{1}{f_{jk}}\frac{d f_{jk}}{dt}\frac{d z_j}{dt}+f_{kj}f_{jk}z_j.$$
using the first relation $z_k=-\frac{1}{f_{jk}}\frac{d z_j}{dt}$ and that if $k=3-j$ then $3-k=j$.
If we compare for example
$$\frac{d^2 y_1}{dt^2}=\frac{1}{f_{12}}\frac{d f_{12}}{dt}\frac{d y_1}{dt}+f_{12}f_{21}y_1$$
$$\frac{d^2 z_1}{dt^2}=+\frac{1}{f_{12}}\frac{d f_{12}}{dt}\frac{d z_1}{dt}+f_{21}f_{12}z_1.$$
unless I've made a mistake with say the chain rule etc, that $y_j$ and $z_j$ satisfy the same differential equation, and since it's linear and homogenous, then $y_j\pm z_j$ should be solutions. That is $y_j(t)\pm y_j(-t)$ is a solution.
It's worth double checking this, but it's an idea anyway.
Edit: If that holds up, and we can choose $y_1$ and $y_2$ even or odd for the second order equation, then in order to satisfy the first pair of equations
$$\frac{d y_1}{dt}=f_{12}y_2 \\ \frac{d y_2}{dt}=f_{21}y_1$$ then one can work out some consistency conditions. Using the properties of producting even and odd functions (even by even is even etc)  and that differentiating (even'=odd etc) , say $+$ means even function and $-$ means odd
\begin{array}{|c|c|c|c|}
\hline
f_{12}& f_{21} & y_1 & y_2 \\ \hline
+ & +& \pm & \mp\\ \hline
+ & - &  &\\ \hline
- & + & &\\ \hline
- & - & \pm& \mp\\ \hline
\end{array}
Again double checking needed it looks like with $f_{12}$ and $f_{21}$ both odd or both even, then one can in theory have a pair of solutions $(y_1,y_2)$ with one an odd function and one an even function for consistency.
