In @MarkusScheuer's excellent answer on Mathematical Explanation for the Repertoire Method the introduction gives a concrete example with two provided repertoire items, $x_n$ and $y_n$ which are then linearly combined to find the solution to $z_n$. Quoting that section below:
To make it more concrete, let $a_n = 3$ and $b_n = 5n^2 + 1$. Assuming we know the solutions of $x_n$ and $y_n$ of the recurrences
$$ \begin{align*} x_0&=3&y_0&=1\\ x_n&=3+x_{n-1},\quad n>0&y_n&=5n^2+1+y_{n-1},\quad n>0 \end{align*} $$
then we also know by linearity that the solution of the recurrence
$$ \begin{align*} z_0&=7\\ z_n&=2n^2+7+z_{n-1} \end{align*} $$
is
$$ \begin{align*} z_n=\frac{11}{5}x_n+\frac{2}{5}y_n \end{align*} $$
To find the eventual closed form cooefficient for the $n^2$ variable (say $\beta$) I think we do the following
$$ \begin{align} 2 &= \beta 5 \\ \frac25 &= \beta \end{align} $$
to use with $y_n$. However I have not been able to figure out how to arrive at $\frac{11}{5}$. I am not sure how to "map" the constant in recurrence $x_n = 3 + x_{n-1}$ to the $7$ in recurrence $z_n$ as well as take into account the $1$ in $y_n = 5n^2 + 1 + y_{n-1}$. I suspect I'm overthinking this.
