Method of Annihilators Tedium... One of the exam preparation questions for MIT's online Honors Differential Equations course asks for a general solution of 
\begin{align}
(D^2 - 1)^4(D^3 + 1)^5y = 3e^t
\end{align}
The fact that the equation is written in terms of differential operators makes me think that I'm to use the method of annihilators; indeed, it's easy enough to see that $D-1$ annihilates $3e^t$.  So we have the homogeneous ODE
\begin{align}
(D^2 - 1)^4(D^3 + 1)^5(D - 1) & = 0\Rightarrow\\
(D + 1)^9(D-1)^5\left(D - \frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^5\left(D + \frac{1}{2} - i\frac{\sqrt{3}}{2}\right)^5 & = 0
\end{align}
which means we have (the absurdly unwieldy, at least to me) basis of solutions
\begin{align}
 \lbrace e^{-t},te^{-t},\ldots, t^8e^{-t}, e^t, te^t, \ldots, t^4e^t, e^{\frac{t}{2}}\cos\left(\frac{x\sqrt{3}}{2}\right), te^{\frac{t}{2}}\cos\left(\frac{t\sqrt{3}}{2}\right), \ldots, t^4e^{\frac{t}{2}}\cos\left(\frac{t\sqrt{3}}{2}\right),\\ e^{\frac{t}{2}}\sin\left(\frac{t\sqrt{3}}{2}\right), te^{\frac{t}{2}}\sin\left(\frac{t\sqrt{3}}{2}\right), t^4e^{\frac{t}{2}}\sin\left(\frac{t\sqrt{3}}{2}\right)\rbrace
\end{align}
My question is this:  do I now really need to go through what appears to be an insane amount of differentiation/algebra to get the particular solution?  Or is there some trick I'm not seeing/fundamental fact I'm missing?

If $\lambda$ is a root of the polynomial $p$ with order $m$, then $t^i \exp(\lambda t)$ for $0 \le i \le m-1$ are solutions of the homogeneous equation $p(D) u = 0$, while $p(D) t^m \exp(\lambda) t$ will be a scalar multiple of $\exp(\lambda) t$.  You just need to figure out which scalar it is.
In this case your polynomial is $p(D) = (D^3+1)^5(D^2-1)^4 = (D^3+1)^5(D+1)^4(D-1)^4$.
I suggest you start with $(D-1) t^4 \exp(t) = 4 t^3 \exp(t)$, ...., 
$(D-1)^4 t^4 \exp(t) = 24 \exp(t)$, and then 
$p(D) t^4 \exp(t) = 2^9 \cdot 24 \exp(t)$. 
