Example 3.12 Solve the recurrence relation:
$$
\left\{\begin{array}{l}
a_{n}=8 a_{n-1}-22 a_{n-2}+24 a_{n-3}-9 a_{n-4} \quad(n \geqslant 4), \\
a_{0}=-1, a_{1}=-3, a_{2}=-5, a_{3}=5 .
\end{array}\right.
$$
The characteristic equation of the given recurrence relation is
$$
x^{4}-8 x^{3}+22 x^{2}-24 x+9=0 \text {, }
$$

The characteristic roots are $x_{1}=x_{2}=1, x_{3}=x_{4}=3$, so
$$
a_{n}=c_{1}+c_{2} n+c_{3} \cdot 3^{n}+c_{4} \cdot n 3^{n} \text {, }
$$

where $c_{1}, c_{2}, c_{3}, c_{4}$ are undetermined constants. From the initial conditions, we get
$$
\left\{\begin{array}{l}
c_{1}+c_{3}=-1, \\
c_{1}+c_{2}+3 c_{3}+3 c_{4}=-3, \\
c_{1}+2 c_{2}+9 c_{3}+18 c_{4}=-5, \\
c_{1}+3 c_{2}+27 c_{3}+81 c_{4}=5 .
\end{array}\right.
$$

Solving this system of equations yields $c_{1}=2, c_{2}=1, c_{3}=-3, c_{4}=1$, hence
$$
a_{n}=2+n-3^{n+1}+n \cdot 3^{n} \quad(n=0,1,2, \cdots) .
$$