Example 5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}-2 a_{n}$ $-3 a_{n-1}=0, a_{1}=1, a_{2}=5$, find its general term formula.
Solve $a_{n+1}-a_{n}=b_{n}$, then we have $b_{n}=-3 b_{n-1}$, which yields $b_{n}=4(-3)^{n-1}$. Thus,
$$
\begin{aligned}
a_{n}= & 1+4 \cdot \frac{1-(-3)^{n-1}}{1+3} \\
= & 2-(-3)^{n-1}
\end{aligned}
$$