If I want to find the annihilator for $2t + e^{-t}$ on the right hand side of an ODE.
I know that one part of it must be $(D + 1)$ to correspond to the $-1$ root for $e^{-t}$.
However, for $2t$:
How do I know whether to interpret it as $((0)(e^0t) + 2(te^0t))$ i.e. multiplicity two of root $0$ with coefficients of $0$ and $2$, which would yield $D^2$ as an annihilator
OR
Multiplicity one of root $0$, with a coefficient of $2t$. i.e. $(2t)(e^0t)$ which would yield $D$ as an annihilator.
Thank you!
If you view $D$ as the derivative operator, then $D(2t)=2,$ so $D^2(2t)=0,$ and your annihilator is as follows
$$\underbrace{D^2(D+1)}_{\text{annihilator}}(2t+e^{-t})=0.$$
Seeing $0$ as a root means that you have a generalized eigenvalue of $0,$ and applying $D$ once more than the degree of your polynomial in question will yield the desired annihilator.