Example: Given $a_{1}=a_{2}=1, a_{n+1}=a_{n}+a_{n-1}(n \geqslant 2)$, find the general term formula of the sequence $\left\{a_{n}\right\}$.
Let $a_{n+1}-\alpha a_{n}=\beta\left(a_{n}-\alpha a_{n-1}\right)$, then $a_{n+1}=(\alpha+\beta) a_{n}-\alpha \beta a_{n-1}$. Comparing with the original recurrence relation, we have $\alpha+\beta=1, \alpha \beta=-1$, thus, $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Therefore, $\alpha, \beta=\frac{1}{2}(1 \pm \sqrt{5})$. Hence, $\left\{a_{n+1}-\alpha a_{n}\right\}$ and $\left\{a_{n+1}-\beta a_{n}\right\}$ are geometric sequences with common ratios $\beta$ and $\alpha$ respectively. So, $a_{n+1}-\alpha a_{n}=\beta^{n-1}(1-\alpha \cdot 1)=$ $\beta^{n}$. Similarly, $a_{n+1}-\beta a_{n}=\alpha^{n}$. Subtracting these two equations, we get $a_{n}=\left(\alpha^{n}-\beta^{n}\right) /(\alpha-\beta)$.
Therefore, $a_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$.
In general, for a sequence determined by the recurrence relation $a_{n}=p a_{n-1}+q a_{n-2}(q \neq 0, n=2,3, \cdots)$, we first find the roots $x_{1}, x_{2}$ (called characteristic roots) of the equation $x^{2}-p x-q=0$ (called the characteristic equation).
(1) When $x_{1} \neq x_{2}$, $a_{n}=\alpha_{1} x_{1}^{n}+\alpha_{2} x_{2}^{n}$
(2) When $x_{1}=x_{2}$, $a_{n}=\left(\beta_{1}+n \beta_{2}\right) x_{1}^{n}$, where $\alpha_{i}, \beta_{i}(i=1,2)$ are constants determined by the initial values.
The method of finding the general term of a recursive sequence using the above conclusions is called the characteristic root method.