The case of any two $\lambda_k$ being equal is trivially true as both sides are necessarily zero.
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@@ -113,14 +118,14 @@ The special case is equivalent to $M=1$ and $\lambda_1 = 0$.
As such, the set $E =\set{\Poly{j}}_{j=0}^{m-1}$.
The following lemma presents the Wronskians for such polynomials:
\begin{lemma}[Wronskians of polynomials]
\begin{lem}[Wronskians of polynomials]
\begin{align*}
(i) &&\Wpoly{k=0}{m}& = 1 \\
(ii) &&\Wpoly{k=0,k \neq j}{m}& = \Wpoly{k=1}{m-j}\\
(iii) &&\Wpoly{k=1}{m}& = \Poly{m}
\end{align*}
\label{lem:poly}
\end{lemma}
\end{lem}
\begin{proof}
\begin{description}
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@@ -218,12 +223,12 @@ The case for one root with multiplicity $m$, represented by $M=1$, is a generali
The set $E = E_1=\set{\Poly{k} e^{\lambda_1 x}}_{k=1}^m$ which is the set $E$ from the previous case multiplied by $e^{\lambda_1 x}$.
The following lemma then makes the generalization simple.
\begin{lemma}
\begin{lem}
\begin{equation*}
W(\{ f_k g \}_{k=1}^m ; x) = g^m W(\{ f_k \}_{k=1}^m ; x )
\end{equation*}
\label{lem:group}
\end{lemma}
\end{lem}
\begin{proof}
It is trivially true for $m=1$.
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@@ -436,4 +441,78 @@ One may write out equation (\ref{eq:Wronskian system}) with this in mind:
where $\tilde{I}_k $ indicates the selection of those rows corresponding to $\lambda_k$.
Note that all $F_k^{-1}(x)$ are principal submatrices of $F_{k^*}^{-1}$, where $m_{k^*}\geq m_k$ for all $k=1,...,M$.
\section{Other attempts}
\newcommand{\Lcal}{\mathcal{L}}
\newcommand{\ddx}{\frac{d}{dx}}
Let $\hat{\Lcal}$ be the operator for the set $\set{\Poly{k} e^{\lambda_j x}}_{j \neq q}$ and $\tilde{\Lcal}$ the one for $\set{\Poly{k} e^{\lambda_q x}}_{k \neq n}$, then
\item$E \to\Lcal$ such that $\forall f \in E$$\Lcal f =0$.
\item$\cup_{i \neq k} E_i \to\tilde{\Lcal}$ in the same way: $\tilde{\Lcal} f = f^{(m-m_k)}(x)+\sum_{i \neq k}\lambda_i^{m_i} f^{(m - m_k -1)}(x)+\dots$.