Example 5. Let the sequence be $u_{1}=2, u_{n+1}=2 u_{n} +3 \cdot 2^{n}$, find its general term.
In this example, since the base of the additional term and the coefficient of $x_{n}$ are both 2, the transformation used in Example 4 is effective. At this point, we can let
$$
\begin{array}{l}
u_{n}=v_{n}+a \cdot a \cdot 2^{n}, \text { thus we have } \\
v_{n+1}+a \cdot(n+1) \cdot 2^{n+1} \\
=2\left(v_{n}+a n \cdot 2^{n}\right)+3 \cdot 2^{n} .
\end{array}
$$

By comparing the coefficients of $2^{\mathrm{n}}$, we find that when $\alpha=\frac{3}{2}$, the above equation becomes $v_{n+1}=2 v_{n}, v_{1}=-1$. Therefore, $v_{n}=-2^{n-1}, u_{n}=(3 n-1) \cdot 2^{n-1}$.

Explanation: 1 From the previous two examples, we can deduce that for the recurrence relation $u_{n+1}=a u_{n}+b a^{n}+c_{1} q_{1}^{n}+c_{2} q_{2}^{n}+\cdots+c_{m} q_{m}^{n}$, it can be transformed by the transformation $u_{n}=v_{n}+a \cdot n \cdot a^{n}+\beta_{1} q_{1}^{n}+\beta_{2} q_{2}^{n}+\beta_{m} q_{m}^{n}$ (where $a, b, c_{1}, q_{1}, \cdots, m$ are constants, $m$ is a natural number, and $\alpha, \beta$ are undetermined constants) into the recurrence relation of a geometric sequence $i_{n+1}=x_{i}=$.
$$
\begin{aligned}
2^{\circ} \text { From } \sin x & =\frac{e^{i x}-e^{-i x}}{2 i}, \\
\cos x & =\frac{e^{i x}+e^{-i x}}{2} \text {, we know that sine and cosine are linear combinations of exponential }
\end{aligned}
$$

functions, so for the sequence $\boldsymbol{u}_{n+1}=$ $a u_{n}+b \sin n \theta+c \cos n \theta$, it can be transformed by the transformation $u_{n}=v_{n}+\alpha \sin n \theta+\beta \cos n \theta$ into $v_{n+1}=a v_{n}$.