Example 3.15 Solve the recurrence relation
$$
\left\{\begin{array}{l}
a_{n}=4 a_{n-1}-3 a_{n-2}+2^{n} \quad(n \geqslant 2), \\
a_{0}=2, \quad a_{1}=8 .
\end{array}\right.
$$
The characteristic equation of the recurrence relation
$$
a_{n}=4 a_{n-1}-3 a_{n-2} \quad(n \geqslant 2)
$$

is $x^{2}-4 x+3=0$, with characteristic roots $x_{1}=1, x_{2}=3$. Therefore, the general solution is $a_{n}=c_{1}+c_{2} \cdot 3^{n}$. Since 2 is not a characteristic root of equation (3.11), the recurrence relation
$$
a_{n}=4 a_{n-1}-3 a_{n-2}+2^{n} \quad(n \geqslant 2)
$$

has a particular solution $a_{n}=A \cdot 2^{n}$, where $A$ is a constant to be determined. Substituting into equation (3.12) yields
$$
\begin{array}{c}
A \cdot 2^{n}=4 A \cdot 2^{n-1}-3 A \cdot 2^{n-2}+2^{n}, \\
A=-4 .
\end{array}
$$

Thus,
$$
a_{n}=c_{1}+c_{2} \cdot 3^{n}-4 \cdot 2^{n},
$$

where $c_{1}, c_{2}$ are constants to be determined. From the initial conditions, we have
$$
\left\{\begin{array}{l}
c_{1}+c_{2}-4=2, \\
c_{1}+3 c_{2}-8=8 .
\end{array}\right.
$$

Solving these, we get $c_{1}=1, c_{2}=5$. Therefore,
$$
a_{n}=1+5 \cdot 3^{n}-4 \cdot 2^{n} \quad(n=0,1,2 \cdots) .
$$