Example 1 (2004 Chongqing College Entrance Examination - Liberal Arts) Given $a_{1}=1, a_{2}=\frac{5}{3}, a_{n+2}=\frac{5}{3} a_{n+1}-\frac{2}{3} a_{n}\left(n \in \mathbf{N}^{*}\right)$, find the general term formula $a_{n}$ of the sequence $\left\{a_{n}\right\}$.
Solution: Consider its characteristic equation $x^{2}=\frac{5}{3} x-\frac{2}{3}$, solving yields $x_{1}=\frac{2}{3}, x_{2}=1$.
Therefore, we can assume $a_{n}=A \cdot\left(\frac{2}{3}\right)^{n}+B \cdot 1^{n}\left(n \in \mathbf{N}^{*}\right)$.
From the values of $a_{1}, a_{2}$, we get
$\left\{\begin{array}{l}a_{1}=A \cdot\left(\frac{2}{3}\right)^{1}+B \cdot 1=1 \\ a_{2}=A \cdot\left(\frac{2}{3}\right)^{2}+B \cdot 1^{2}=\frac{5}{3}\end{array}\right.$, solving yields $A=-3, B=3$.
Thus, $a_{n}=\left(\frac{2}{3}\right)^{n} \cdot(-3)+3$.
Analysis: This problem is a typical $a_{n+1}=\alpha a_{n}+\beta u_{n+1}$ type. When solving such problems, the characteristic root method can be used, and the general method is as follows:
1) List the characteristic equation from the recurrence relation and find its characteristic roots $x_{1}, x_{2}$;
2) Assume the general term of the sequence $a_{n}=A \cdot\left(x_{1}\right)^{n}+B \cdot\left(x_{2}\right)^{n}$;
3) Solve for $A, B$ using the first few terms of the sequence, thus obtaining the general term formula $a_{n}$.
For the generalization of this type of problem, i.e., $a_{n+2}=\alpha a_{n}+\beta a_{n-1}+\gamma$ type problems, we can first transform it into $a_{n-2}+$ $\frac{\gamma}{\alpha+\beta-1}=\alpha\left(a_{n}+\frac{\gamma}{\alpha+\beta-1}\right)+\beta\left(a_{n+1}+\frac{\gamma}{\alpha+\beta-1}\right)$, then substitute $b_{n}$ for $a_{n}+\frac{\gamma}{\alpha+\beta-1}$, transforming it into $b_{n+2}=\alpha b_{n}+\beta b_{n-1}$, at which point the characteristic root method can be used to solve. Note: When $\alpha+\beta=1$, the above method cannot be used. In this case, it can be transformed into $a_{n+2}-a_{n-1}-\frac{\gamma}{\beta+1}=-\beta\left(a_{n+1}-a_{n}-\frac{\gamma}{\beta+1}\right)$, and it is easy to see that $\left\{a_{n+2}-a_{n+1}-\frac{\gamma}{\beta+1}\right\}$ is a geometric sequence with a common ratio of $-\beta$, making $a_{n}$ easy to find.