Theorem from Birkhoff and Rota -- method of undetermined coefficients(?) In Section 3.5 of Birkhoff and Rota's text Ordinary Differential Equations, 4th Ed. the following theorem is stated:
Given a polynomial $r(t)$, the differential equation $L[u] = e^{\lambda}r(t)$ has a particular solution of the form $e^{\lambda t}q(t)$, where $q(t)$ is also a polynomial.  The degree of $q(t)$ equals that of $r(t)$, unless $\lambda = \lambda_j$ is a root of the characteristic polynomial $p_L(\lambda) = \sum(\lambda - \lambda_j)^{k_j}$ [sic -- clearly the authors intended $\prod$ in place of $\sum$] of $L$.  If $\lambda = \lambda_j$ is a $k$-fold root of $p_L(\lambda)$, then the degree of $q(t)$ exceeds that of $r(t)$ by $k$.
To prove this, we are instructed to apply a pair of easily proven lemmas repeatedly to each factor of the operator $L = p_L(D) = \prod(D - \lambda_j)^{k_j}$.  These lemmas are:
(1)  If $r(t)$ is a polynomial of degree $s$, then $(D - \lambda)[u] = e^{\lambda t}r(t)$ has a solution of the form $u = e^{\lambda t}q(t)$, where $q(t)$ is a polynomial of degree $s + 1$.
(2)  If $r(t)$ is a polynomial of degree $s$ and $\lambda \neq \lambda_1$, then
$(D - \lambda_1)[u] = e^{\lambda t}r(t)$ has a solution of the form $u = e^{\lambda t}q(t)$, where $q(t)$ is a polynomial of degree $s$.
I'm afraid I don't quite see it.  For assume $\lambda$ is not a root of $p_L$, and assume (to keep things simple) that $p_L$ only has roots of multiplicity $1$.  Then Lemma 2 certainly guarantees that for each factor $D - \lambda_j$ of $p_L(D)$, the DE $(D - \lambda_j)[u] = e^{\lambda t}r(t)$ has a solution of the form $e^{\lambda t}q_j(t)$ -- but I do not see how we know that for each of the factors $(D - \lambda_j)$ of $p_L(D)$ the polynomials $q_j(t)$ will be the same.  (Indeed, it would seem that each $q_j$ would necessarily vary with $\lambda_j$.)  Or are we supposed to be somehow constructing a 'master' $q(t)$ of some sort from these various $q_j(t)$?  In which case, are the 'mini' DEs we are to use to obtain these $q_j(t)$ not of the form $(D - \lambda_j)[u] = e^{\lambda t}r(t)$ but instead of the form $(D - \lambda_j)[u] = e^{\lambda t}r_j(t)$, with $r_j(t)$ necessarily of smaller degree than $r(t)$?  (Moreover--and I recognize that I may have chased a wild goose (or red herring, or similar metaphorical beastie) a lot farther than is healthy by now -- how would one determine such $r_j$ in the first place?)
Any help untangling this proof would be most appreciated. 
 A: The problem here is that I was trying to find a specific $q(t)$, rather than merely demonstrating that a suitable $q(t)$ exists.  In other words, I was mistaking an existence proof for a construction proof.  (Lesson:  don't post when you're exhausted.)
So there are two cases:


*

*$\lambda$ is not a root of $p_L$.  Write $p_L(D)$ as $(D - \lambda_1) \cdots (D - \lambda_n)$ where all factors have a multiplicity of $1$ (i.e. some may be repeated.)  Now apply Lemma 2 iteratively over all $n$ factors of $p_L(D)$.  Since $\lambda \neq \lambda_i$ for all $i$, we successively demonstrate the existence of polynomials $q_i(t)$ of degree $s$ such that $(D - \lambda_1)\cdots(D-\lambda_i)(e^{\lambda t}q_i(t)) = e^{\lambda t}r(t)$.  When we reach $q_n(t)$, we have a polynomial of degree $s$ such that $p_L(D)(e^{\lambda t}q_n(t)) = L[e^{\lambda t}q_n(t)] = e^{\lambda t}r(t)$, as desired.

*$\lambda$ is a root of $p_L$; assume without loss of generality that $\lambda = \lambda_1$.  Write $p_L(D)$ as $(D - \lambda_1)(D- \lambda_1)\cdots (D-\lambda_1)(D - \lambda_2) \cdots (D - \lambda_n)$, where each factor is repeated $k_i$ times, where $k_i$ is the multiplicity of the root $\lambda_i$.  Now apply Lemma 1 $k_1$ times to $(D - \lambda_1)$; we obtain a polynomial $q_1$ of degree $s - k_1$ such that $(D - \lambda_1)^{k_1}(e^{\lambda t}q_1(t)) = e^{\lambda t}r(t)$.  Finally, apply Lemma 2 to each remaining factor of $p_L(D)$; we eventually obtain a polynomial $q_n(t)$ of degree $s - k_1$ such that $p_L(D)(e^{\lambda t}q_n(t) = L[e^{\lambda t}q_n(t)] = e^{\lambda t}r(t)$, which is what we want.

