Example 7 (1993 National High School League Question) The sequence of positive numbers $\left\{a_{n}\right\}$ satisfies $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}$, and $a_{0}=a_{1}=1$, find the general formula for $a_{n}$.
Analysis: Removing the square root is the key to this problem. Through attempts, it is not difficult to find that dividing both sides by $\sqrt{a_{n-1} a_{n-2}}$ and then using the substitution method can transform it into a linear recurrence sequence problem to solve.
Solution: Clearly, $\sqrt{a_{n-1} a_{n-2}}$ is non-zero. Dividing both sides of the original equation by $\sqrt{a_{n-1} a_{n-2}}$ yields $\sqrt{\frac{a_{n}}{a_{n-1}}}-1=2 \sqrt{\frac{a_{n-1}-1}{a_{n-2}}}$.
Let $b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}}(n \geqslant 1)$, then $b_{1}=1$, and $b_{n}=2 b_{n-1}+1(n \geqslant 3)$.
Assume $b_{n}=A \cdot 2^{n}+B \cdots(1)$
Solving for $b_{1}, b_{2}$ gives $\left\{\begin{array}{l}A=1 \\ B=-1\end{array}\right.$.
Thus, $b_{n}=2^{n}-1$, and therefore $\frac{a_{n}}{a_{n-1}}=b_{n}^{2}=\left(2^{n}-1\right)^{2}$.
Hence, $a_{n}=\frac{a_{n}}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_{1}}{a_{0}} \cdot a_{0}=\prod_{i=1}^{n}\left(2^{n}-1\right)^{2}(n \in \mathbf{N})$.
Comment: The transformation at (1) is applicable to the form $x_{n}=\alpha x_{n-1}+\beta$. This is because $x_{n}=\alpha x_{n-1}+\beta$, and $x_{n-1}=\alpha x_{n-2}+\beta$. Subtracting the two equations gives $x_{n}-(\alpha+1) \alpha_{n-1}+\alpha x_{n-2}=0$. By the characteristic root method, the roots of $x^{2}-(\alpha+1) x+\alpha=0$ are $x_{1}=\alpha, x_{2}=1$. Therefore, we can set $x_{n}=A \cdot(\alpha)^{n}+B$.
Additionally, when solving sequences of the form $x_{n}=\alpha x_{n-1}+\beta$, they can be simplified and solved using the substitution method to transform them into geometric sequences.