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?
 A: 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 
$p(D) t^4 \exp(t) = 2^9 \cdot 24 \exp(t)$. 
A: You can reduce the tedium. The only solution of
$$
             (D-1)(D^2-1)^4(D^3+1)^5y = 0 \\
             \mbox{ or } (D-1)^5(D+1)^4(D^3+1)^5y = 0
$$
that is not solutions of
$$
             (D^2-1)^4(D^3+1)^5y=0 \\
             \mbox{ or } (D-1)^4(D+1)^4(D^3+1)^5y = 0
$$
is a single term of the form
$$
                      Ct^4e^{t}
$$
So there's only one term you need to plug into the original equation, because all other terms are annihilated, and that gives the following equation:
$$
            (D+1)^4(D^3+1)^5(D-1)^4(Ct^4e^{t})=3e^t.
$$
The reason I moved $(D-1)^4$ to the far right is that this term acts in a very predictable way on $Ct^4e^{t}$ because $(D-1)f = e^{t}D(e^{-t}f)$. Hence,
$$
          (D-1)^{4}f = e^{t}D^4(e^{-t}f) \\
          (D-1)^{4}(Ct^4e^{t})=e^{t}D^4(Ct^4)=Ce^{t}(4)(3)(2)(1)=24Ce^{t}.
$$
So, you want
$$
      (D+1)^4(D^3+1)^5(24Ce^{t})=3e^t \\
      24C(1+1)^4(1^3+1)^5 e^{t}=3e^{t} \\
         C = \frac{3}{24}\frac{1}{2^9}
$$
The general solution of the original equation is the specific solution $Ct^4e^t$ plus the solution of the homogeneous equation
$$
         (D-1)^5(D+1)^4(D+1)^5(D^2-D+1)^5y=0 \\
         (D-1)^5(D+1)^9((D-1/2)^2+3/4)^5y=0.
$$
I assume you can handle the homogeneous. It looks like you already have.
