11. (This question is worth 20 points) The sequence $\left\{x_{n}\right\}$ satisfies $x_{0}=0$ and $x_{n+1}=3 x_{n}+\sqrt{8 x_{n}^{2}+1}$, find the general term formula.
From $x_{n+1}=3 x_{n}+\sqrt{8 x_{n}^{2}+1}$, we get $\left(x_{n+1}-x_{n}\right)^{2}=8 x_{n}^{2}+1$, which means
$$
x_{n+1}^{2}-6 x_{n+1} x_{n}+x_{n}^{2}-1=0 \text {, }
$$

By replacing the subscript $n$ with $n-1$ in the above equation, we get $\left(x_{n+1}-x_{n}\right)^{2}=8 x_{n}^{2}+1$, which means
$$
x_{n-1}^{2}-6 x_{n+1} x_{n}+x_{n}^{2}-1=0,
$$

From these two equations, we know that $x_{n+1}, x_{n-1}$ are the two roots of the quadratic equation $t^{2}-6 x_{n} t+x_{n}^{2}-1=0$. By Vieta's formulas, we have $x_{n+1}+x_{n-1}=6 x_{n}$. This is a second-order linear recurrence sequence, whose characteristic equation is $\lambda^{2}-6 \lambda+1=0$, with characteristic roots $\lambda_{1}=3+2 \sqrt{2}, \lambda_{2}=3-2 \sqrt{2}$. ….. 10 points Therefore, we can assume its general term to be $x_{n}=C_{1}(3+2 \sqrt{2})^{n}+C_{2}(3-2 \sqrt{2})^{n}$. From $x_{0}=0$ and $x_{n+1}=x_{n}+\sqrt{8 x_{n}^{2}+1}$, we get $x_{1}=1$, so $C_{1}$ and $C_{2}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
C_{1}+C_{2}=0 \\
(3+2 \sqrt{2}) C_{1}+(3-2 \sqrt{2}) C_{2}=1 .
\end{array}\right.
$$

Solving this, we get $C_{1}=\frac{\sqrt{2}}{8}, C_{2}=-\frac{\sqrt{2}}{8}, x_{n}=\frac{\sqrt{2}}{8}(3+2 \sqrt{2})^{n}-\frac{\sqrt{2}}{8}(3-2 \sqrt{2})^{n} \cdots \cdots 20$ points