Example 8 Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0$, and $a_{n}-2 a_{n-1}=n^{2}-3 \ (n=2,3, \cdots)$, find $a_{n}$.
Solve: From $a_{n}-2 a_{n-1}=n^{2}-3$ we get $a_{n-1}-2 a_{n-2}=(n-1)^{2}-3=n^{2}-2 n-2$, thus $a_{n}-3 a_{n-1}+2 a_{n-2}=2 n-1$.
From $a_{n-1}-3 a_{n-2}+2 a_{n-3}=2(n-1)-1=2 n-3$ we get $a_{n}-4 a_{n-1}+5 a_{n-2}-2 a_{n-3}=2$.
From $a_{n-1}-4 a_{n-2}+5 a_{n-3}-2 a_{n-4}=2$ we get
$$
a_{n}-5 a_{n-1}+9 a_{n-2}-7 a_{_{3}}+2 a_{n-4}=0 .
$$

Thus, its characteristic equation is $x^{4}-5 x^{3}+9 x^{2}-7 x+2=0$, whose roots are $x_{1}=x_{2}=x_{3}=1, x_{4}=2$. Therefore, the general term can be set as $a_{n}=\left(c_{1}+c_{2} n+c_{3} n^{2}\right) \cdot 1^{n}+c_{4} \cdot 2^{n}$.
From $a_{1}=0, a_{n}-2 a_{n-1}=n^{2}-3$ we have $a_{2}=1, a_{3}=8, a_{4}=29$.
Thus, we have $\left\{\begin{array}{l}c_{1}+c_{2}+c_{3}+2 c_{4}=0, \\ c_{1}+2 c_{2}+4 c_{3}+4 c_{4}=1, \\ c_{1}+3 c_{2}+9 c_{3}+8 c_{4}=8, \\ c_{1}+4 c_{2}+16 c_{3}+16 c_{4}=29,\end{array} \quad\right.$ which is $\left\{\begin{array}{l}c_{1}=-1, \\ c_{2}=-4, \\ c_{3}=-3, \\ c_{4}=4 .\end{array}\right.$
Therefore, the general term is $a_{n}=-n^{2}-4 n-3+4 \cdot 2^{n}=2^{n+2}-n^{2}-4 n-3$.