Example 10. Let the sequence be $u_{1}=3$, $s_{n+1}=\frac{2 u_{n}+3}{u_{n}+4}$, find its general term.
Let $u_{n}=v_{n}+\alpha$, then the original equation becomes
$$
v_{n+1}=\frac{(2-\alpha) v_{n}-\left(\alpha^{2}+2 \alpha-3\right)}{v_{n}+\alpha+4}
$$

When $a^{2}+2 a-3=0$, it reduces to the case of Example 9. For this, take $a=1$, and let $v_{1}=2$,
$$
v_{n+1}=\frac{v_{n}}{v_{n}+5} .
$$

Let $\mu_{n}=\frac{1}{v_{n}}$, then $\mu_{1}=\frac{1}{2}, w_{n+1}$
$=5 w_{n}+1$, and using the method of Example 1, we can solve to get:
$$
\begin{array}{l}
w_{n}=\frac{1}{4}\left(3 \cdot 5^{n-1}-1\right), v_{n}=\frac{4}{3 \cdot 5^{n-1}+1}, \\
u_{\mathrm{n}}=\frac{3 \cdot 5_{n}+3}{3 \cdot 5_{n}-1} .
\end{array}
$$

Explanation: $1^{\circ}$ This method applies to the recursive sequence $u_{\mathrm{n}+1}=\frac{a u_{n}+b}{c u_{n}+d}$. The equation $\alpha^{2}+2 \alpha-3=0$ in the example is called the characteristic equation, which is the equation satisfied by the fixed point $x=\frac{a x+b}{c x+d}$ of the function $f(x)=\frac{a x+b}{c x+d}$: $c x^{2}+(d-a) x-b=0$.
$2^{\circ}$ The characteristic equation has another solution $\alpha=-3$, i.e., we can let $u_{n}=v_{n}-3$, and the results obtained are completely consistent.