First, let $$d = h(x) \frac{d}{dx} + k(x) Id$$ be a smooth differential operator, with $h$ nowhere zero. Setting $x'(t) = h(x(t))$ (with solution given by $x(0)=c$, for some $c$, say) as our change of variables, we get $$ \frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt} = h(x(t)) \frac{df}{dx},$$ so we can rewrite the operator as $$d = \frac{d}{dt} + k(x(t))Id.$$ This essentially cancels out the leading coefficient function and simplifies studying this operator.
Now I'm interested in doing the same thing, but with the following operator which uses matrices: $$ D = H(x) \frac{d}{dx} + K(x) $$ where $H$ and $K$ are $2 \times 2$ matrices for simplicity, with smooth functions as entries, and where $H$ is invertible for every $x$. Here, the derivative acts on matrices entry-wise. But I'm stuck on showing that this can be reduced to something like $$ D = \frac{d}{dx} + \tilde{K}(x). $$
How does one do this change of variables, and using what tricks? Is it similar to the simple case at the beginning?