A structured differential equation I stumbled into this structured vector-type ODE:
$\dot{x} + \xi x = f$, with initial condition $x(0)=0$,
where $x(t):[0,\ 1]\to\mathbb{R}^n$ denotes the unknown, $\xi(t)$ is a known $n\times n$ skew-symmetric-valued matrix function of the parameter $t$ only and $f(t)$ is a known $\mathbb{R}^n$-valued function of the parameter $t$ only.
I was wondering if anybody knows of any closed-form solution. Thanks!
 A: Let us assume that $n=2$: in that case I claim that there is indeed a closed form. As a preliminary to the proof, I would like to begin with a remark on first-order linear ODE; of course in one dimension the equation
$$
\dot x +a(t) x=0,\quad x(0)=x_0.
$$
has the unique solution $e^{-A(t)} x_0$ where $\dot A=a, A(0)=0$. It is no longer true in $n$ dimensions although you can try the same formula when $a$ is a $n\times n$ matrix: this does not work in general because of non-commutation of the matrix $\dot A$ with $A$, but if that commutation holds true it works. Let us go back to your problem with $n=2$: you have for the homogeneous case with $x$ a $2\times 1$ matrix
$$
\dot x+\alpha(t)\begin{pmatrix}0&1\\
-1&0\end{pmatrix}x=0, \quad x(0)=x_0.
\tag 1\label{1}$$
Now the matrix $A$ defined above is
$$
A(t)=\begin{pmatrix}0&1\\
-1&0\end{pmatrix}\int_0^{t} \alpha(t') dt'
$$
and the commutator $[\dot A,A]=0$ identically. As a consequence, the unique solution of \eqref{1} is indeed given by $e^{-A(t)} x_0$; the variation of constant method is also providing the expected answer for the non-homogeneous problem linked to \eqref{1},
that is
$$
x=e^{-A(t)} x_0+\int_0^t e^{-(A(t)-A(t'))} F(t') dt'
$$
is the unique solution of
$
\dot x +a(t) x=F,\quad x(0)=x_0.
$
You can of course repeat ne varietur this argument in $2n$ dimensions if the matrix
$$
a(t)=\alpha(t)\begin{pmatrix}0&I_n\\
-I_n&0\end{pmatrix}
$$
where $0$ here stands for the ${n\times n}$ zero matrix. However, in three or more dimensions, this is hopeless to expect a closed form in general, but you could keep in mind that the identity $[\dot A,A]\equiv 0$ could save you,
which happens in two dimensions because that identity follows from the skew-symmetry hypothesis.
