I'm at a sticky spot in my research. Namely, I have a particular fact, and it ought to have a short proof, but the only way I know how to show it is long and drawn out, and I don't like it and worry I might have an error. I'm hoping one of y'all will either see a short proof or respond with "all your questions are answered in [link]". And I'm hoping this isn't too close to "homework question".

I have a linear second-order differential operator $\mathcal D$ on $C^\infty( [0,1], \mathbb R^n)$, where $\mathbb R^n$ has its usual metric, and of the following form: $$ \mathcal D = \frac{d^2}{dt^2} + B(t) \frac{d}{dt} + C(t) $$ where $B,C$ are $n\times n$ matrix-valued functions on $[0,1]$, $B(t)$ is antisymmetric for each $t$, and $C(t) - C(t)^{\rm T} = B'(t)$, where $C^{\rm T}$ is the transpose of $C$. I happen to know a lot of solutions to $\mathcal D[f] = 0$. In particular, I have two matrix-valued functions $f\_1(t)$ and $f\_2(t)$, which satisfy $\mathcal D[f\_a] = 0$, and also $f\_1(0) = \delta = f\_2(1)$ and $f\_2(0) = 0 = f\_1(1)$, where $\delta$ is the unit $n\times n$ matrix.

(Incidentally, this implies that the columns of the $f\_a$ are a basis for the space of solutions of $\mathcal D[f]=0$, so that there are no nonzero solutions with $f(0) = 0 = f(1)$. Indeed, any solution with $\mathcal D[f] = 0$, $f(0) = 0$ is determined by the derivative $f'(0)$, so that there is a linear map $\mathbb R^n \to \mathbb R^n$ sending $v$ to the value $f(1)$ where $f'(0) = v$. But $f\_2(1) = \delta$, and so $f\_2'(0)$ is full-rank, and so if $f$ solves the differential equation with $f(0) = 0$, then $f(t) = f\_2(t)\left(f\_2'(0)\right)^{-1}f'(0)$.)

Anyhoo, so my question is this. Let $g\_1(t),g\_2(t)$ be matrix-valued functions such that:
$$ f\_1g\_1 + f\_2g\_2 = 0 \text{ and } f\_1' g\_1 + f\_2' g\_2 = \delta$$
**Prove that $\mathcal D[(g\_a)^{\rm T}] = 0$.**

For example, when $n=1$, $B(t) = 0$ because there are no antisymmetric $1\times 1$ matrices, and then by Abel's formula the determinant of the matrix $\left(\begin{smallmatrix} f\_1 & f\_2 \\\ f\_1' & f\_2' \end{smallmatrix}\right)$ is constant. Therefore, $g\_2$, which is the lower-right corner of the inverse of this matrix, is a constant times $f\_1$, and $g\_1$, which is the upper right-hand-corner of the inverse, is a constant times $f\_2$.