Skew-symmetric, Time Dependent, Linear Ordinary Differential Equations Let
$I \subset \Bbb R \tag 1$
be an open interval, not necessarily bounded, and let $A(t)$ be a continuous matrix function of $t \in I$, taking values in
$M(n, \Bbb R)$; that is
$A(t) \in C^0(I, M(n,, \Bbb R)); \tag 2$
suppose further that $A(t)$ is skew-symmetric for every $t$:
$A^T(t) = -A(t), \tag 3$
and consider the ordinary time-varying linear system
$\dot{\vec x}(t) = A(t) \vec x(t), \tag 4$
where
$\vec x(t) \in C^1(I, \Bbb R^n). \tag 5$
It is well-known, and easy to see, that when $A$ is a constant matrix the solutions to (4) are given by
$\vec x(t) = e^{A(t - t_0)} \vec x(t_0), \tag 6$
where the matrix exponential is orthogonal; indeed we have
$(e^{A(t - t_0)})^T e^{A(t - t_0)} = e^{A^T(t - t_0)} e^{A(t - t_0)} = e^{-A(t - t_0)}e^{A(t - t_0)} = I. \tag 7$
The fact that $e^{A(t - t_0)}$ is orthogonal leads to
$\langle \vec x(t), \vec y(t) \rangle = \langle e^{A(t - t_0)} \vec x(t_0), e^{A(t - t_0)} \vec y(t_0) \rangle = \langle \vec x(t_0), \vec y(t_0) \rangle; \tag 8$
that is, the evolution of vectors according to (4) preserves inner products.
We recall that $e^{A(t - t_0)}$ is a fundamental matrix solution of (4), and we observe that it takes the value $I$ at $t = t_0$:
$e^{A(t_0 - t_0)} = e^{A(0)} = I. \tag 9$
The purpose here is to investigate extending this observation to the case in which $A(t)$ is not a constant matrix.  
The Question is then:  given a system of the form (4), with $A(t)$ as in (3), show that a fundamental solution matrix $X(t, t_0)$ of the system (4) with
$X(t_0, t_0) = I \tag{10}$
is orthogonal.  Conversely, show that a system (4) with orthogonal fundamental matrix satisfies (3).
 A: The iterative solution of the celebrated Magnus expansion
for your equation
$$\dot{\vec x}(t) = A(t) \vec x(t), \tag 4$$
uses the Ansatz
$$
\vec x(t) =  e^{\Omega(t,t_0)}  ~ \vec x(t_0)  
$$
so that
$$
A(t)= e^{-\Omega}\frac{d}{dt}e^{\Omega(t)} =  \frac{1 - e^{-\mathrm{ad}_{\Omega}}}{\mathrm{ad}_{\Omega}}\frac{d\Omega}{dt}~~,  
$$
where  ${\mathrm{ad}_{\Omega}} B \equiv  [\Omega, B] $,
$$
A= \dot \Omega - [\Omega, \dot \Omega]/2! + [\Omega,[\Omega,\dot \Omega]]/3! ~ - ~ ...
$$
From (3), however, it follows that $\Omega$ is antisymmetric as well,
$$
A^T=-A=\frac{d}{dt}e^{\Omega^T}   ~e^{-\Omega^T}=- e^{\Omega^T} \frac{d}{dt}e^{-\Omega^T}  =  - e^{-\Omega} \frac{d}{dt}e^{\Omega} .
$$
You may reassure yourself that, indeed, the nested integrals of commutators involved in the solution of $\Omega$ in terms of A  yield antisymmetric results for each term of the iteration.
A: Here is an alternative approach: Writing $x(t) =\Omega_t x(0)$ we have the following ode for $\Omega_t$:
$$ \frac{d}{dt} \Omega_t = A_t \Omega_t, \ \ \ \Omega_t={\bf 1} .$$
Then
$$ \frac{d}{dt} \Omega_t^T \Omega_t = (A \Omega_t)^T \Omega_t + \Omega_t^T (A \Omega_t) =  - \Omega_t^T (A \Omega_t) +\Omega_t^T (A \Omega_t),$$
from which $$ \Omega_t^T \Omega_t = \Omega_0^T \Omega_0 = {\bf 1}, \ t\in {\Bbb R}.$$
