I understand the process for how Eigenvalues are involved in Differential Equations. If you have Differential System of Equations like this
$$ \vec{x}' = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}\vec{x} $$
The solution to that System of Differential Equations is a Linear Combination of e to the power of the eigenvalues times the corresponding eigenvectors.
$$ \vec{x} = C_1e^{2t}\begin{pmatrix} 1 \\ 0 \end{pmatrix} + C_2e^t\begin{pmatrix} -1 \\ 1 \end{pmatrix} $$
However, what I am struggling with figuring out how Generalized Eigenvalues translate to the solutions in Differential Equations. If you take this System of Differential Equations
$$ \vec{x}' = \begin{pmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 6 & 3 & 1 \end{pmatrix}\vec{x} $$
The solution to this Differential Equations is
$$ \vec{x} = C_1e^t\begin{pmatrix} 1 \\ 3t \\ 6t + \frac{9}{2}t^2 \end{pmatrix} + C_2e^t\begin{pmatrix} 0 \\ 1 \\ 3t \end{pmatrix} + C_3e^t\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$
But the eigenvectors of this matrix (including the generalized eigenvectors) are
$$ \vec{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix},\:\: \vec{w}_1 = \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix},\:\: \vec{w}_2 = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} $$
These eigenvectors do share some similarities to the solution to the System of Equations. However, the $$ 6t + \frac{9}{2}t^2$$ term in the solution is one I have no idea how generalized eigenvectors relate. May someone please explain how these eigenvectors translate into Differential Equations?