I struggled understanding the following derivation:
In the beginning, $x$ and $y$ were declared as two functions of a variable $t$. And another function $v$ of $t$ was taken so that, $$\left( D^{N}+\sum_{j=0}^{N-1}a_j D^{j} \right)v=x \qquad\left[\left[D=\frac{d}{dt}\right]\right] \tag1$$
and, $$y=\left(\sum_{i=0}^n b_iD^i\right)v \tag2$$
From these two relations, it was proved that, $$\left(D^{N}+\sum_{j=0}^{N-1}a_j D^{j}\right)y = \left(\sum_{i=0}^n b_iD^i\right)x$$
I don't understand the following proof I came across:
\begin{align} \left(D^{N}+\sum_{j=0}^{N-1}a_j D^{j}\right)y &= \left(\sum_{i=0}^n b_i D^{(i+N)}+\sum_{j=0}^{N-1}a_j \sum_{i=0}^n b_i D^{(i+j)}\right)v \\ &= \left( \sum_{i=0}^n b_i \left( D^{(i+N)} + \sum_{j=0}^{N-1}a_j D^{(i+j)} \right)\right)v && \text{[STEP:1]}\\ &=\left(\sum_{i=0}^n b_iD^i\right)x &&\text{[STEP:2]} \end{align}
I don't understand how STEP-2 follows from STEP-1. Help?
We set $ A = D^N + \sum^{N-1}_{j= 0}a_jD^j$ and $B = \sum^{n}_{i= 0}b_iD^i$.
If now $a_j,b_i$ do not depend on $t$, for any trial function $f$ obviously $ABf = BAf$, i.e. they commute, since this is true for D and any constant ($Da_i = a_iD$) and, of course, for any powers of $D$ ($D^{n}D^{m} = D^{m}D^{n}$).
Now according to your definitions $A\nu = x$ and $B\nu = y$. So we may write $Ay = AB\nu = BA \nu = Bx$. This last relation is shown by the proof in quite a pedestrian way...