Let
$$L(f(t)) = F(s)=N(s)/M(s)$$
be the Laplace Transform of a certain (exponential-bounded) function, $N(s)$ and $M(s)$ being minimal polynomials.
Is it true that the operator $M(D)$ always annihilates $f(t)$ (i.e., $M(D)f(t) = 0$, where $D=d/dt$)?
Does this fact generalize to some more general function types?
I'll illustrate this with several examples:
1) $f(t) = \sum_{k=0}^n c_k t^k, c_k$ are real parameters, $n$ is a natural number. In this case, $F(s)= pol_n(s)/s^{n+1}$, where $pol_n(s)$ is a polynomial in $s$ of degree $n$. $M(s)=s^{n+1}$, so $M(D)=D^{n+1}$, which clearly annihilates $f(t)$.
2) $f(t) = e^{at}, a$ is a real parameter. In this case, $F(s)=1/(s-a)$. $M(s)=s-a$, so $M(D)=D-a$, which clearly annihilates $f(t)$.
3) $f(t) = c_1\sin(wt) + c_2\cos(wt)$, where $w, c_1$ and $c_2$ are real parameters. In this case, $F(s)= (c_1w + c_2s)/(s^2+w^2)$. $M(s)=s^2+w^2$, so $M(D)=D^2+w^2$, which clearly annihilates $f(t)$.
You could actually take advantage of several theorems about the Laplace Transform (e.g. Shift on the $s$-plane) to calculate the annihilator of more complex functions:
4) $f(t) = c_1e^{at}\sin(wt) + c_2e^{at}\cos(wt)$, where $a, w, c_1$ and $c_2$ are real parameters. In this case, $F(s)= (c_1w + c_2(s-a))/((s-a)^2+w^2)$. $M(s)=(s-a)^2+w^2$, so $M(D)=(D-a)^2+w^2$, which annihilates $f(t)$.
5) $f(t) = c_1t\sin(wt) + c_2t\cos(wt)$, where $w, c_1$ and $c_2$ are real parameters. In this case, $F(s)= (2c_1ws + c_2(s^2-w^2))/(s^2+w^2)^2$. $M(s)=(s^2+w^2)^2$, so $M(D)=(D^2+w^2)^2$, which annihilates $f(t)$.
6) $f(t) = c_1t^ne^{at}$, where $a, c_1$ are real parameters and $n$ is a natural number. In this case, $F(s)= c_1n!/(s-a)^{n+1}$. $M(s)=(s-a)^{n+1}$, so $M(D)=(D-a)^{n+1}$, which annihilates $f(t)$.
Finally, you could sum up some of the previous results to get the following:
7) $f(t) = (\sum_{k=0}^n c_k t^k)e^{at}\sin(wt) + (\sum_{k=0}^n d_k t^k)e^{at}\cos(wt)$, where $a, w, c_k$ and $d_k$ are real parameters. In this case, $M(D)=((D-a)^2+w^2)^{n+1}$, which annihilates $f(t)$.
What about $f(t)= \sqrt{t}$, for which $F(s)=\sqrt{\pi}/(2s^{3/2})$?
If $M(x) = \sum_{n=0}^m c_n x^n$ then $M(\frac{d}{dt}) f(t) = \sum_{n=0}^m c_n \frac{d^n}{dt^n} f(t)$ and $$\mathcal{L}[M(\frac{d}{dt}) f(t)](s) = \sum_{n=0}^m c_n \mathcal{L}[\frac{d^n}{dt^n} f(t)](s) = \sum_{n=0}^m c_n s^n F(s)=M(s) F(s)$$ If $M(s) F(s)=\sum_{k=0}^K a_k s^k$ is a polynomial then $$M(\frac{d}{dt}) f(t) =\mathcal{L}^{-1}[\sum_{k=0}^K a_k s^k](t) = \sum_{k=0}^K a_k \delta^{(k)}(t)$$ where $\delta^{(k)}(t)$ is the distributional derivative of the Dirac delta distribution and the $k+1$-th derivative of $ 1_{t > 0}$ or $-1_{t < 0}$ depending on the domain of convergence.
Thus $M(\frac{d}{dt}) f(t) = 0$ only for $t \ne 0$ and around $t= 0$ it is not function.
Does it generalize ? Well this is the convolution theorem for the Fourier/Laplace transform of distributions.