14. Given the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies:
$$
a_{1}=1, a_{2}=1, a_{n+1}=\frac{n^{2} a_{n}^{2}+5}{\left(n^{2}-1\right) a_{n-1}}(n \geqslant 2) \text {. }
$$

Find the general term formula for $a_{n}$.
14. From $\left(n^{2}-1\right) a_{n-1} a_{n+1}=n^{2} a_{n}^{2}+5$,
$$
\left[(n-1)^{2}-1\right] a_{n-2} a_{n}=(n-1)^{2} a_{n-1}^{2}+5 \text {, }
$$

subtracting the two equations gives
$$
\begin{array}{l}
\left(n^{2}-1\right) a_{n-1} a_{n+1}-n(n-2) a_{n-2} a_{n} \\
=n^{2} a_{n}^{2}-(n-1)^{2} a_{n-1}^{2},
\end{array}
$$

which simplifies to $(n-1) a_{n-1}\left[(n+1) a_{n+1}+(n-1) a_{n-1}\right]$
$$
=n a_{n}\left[n a_{n}+(n-2) a_{n-2}\right] \text {. }
$$

Let $b_{n}=n a_{n}(n \geqslant 1)$. Then
$$
b_{n-1}\left(b_{n+1}+b_{n-1}\right)=b_{n}\left(b_{n}+b_{n-2}\right) \text {, }
$$

which implies $\frac{b_{n+1}+b_{n-1}}{b_{n}}=\frac{b_{n}+b_{n-2}}{b_{n-1}}$.
Let $c_{n}=\frac{b_{n}+b_{n-2}}{b_{n-1}}(n \geqslant 3)$.
Given $a_{1}=1, a_{2}=1, a_{3}=3$, we have
$$
b_{1}=1, b_{2}=2, b_{3}=9 \text {. }
$$

Thus, $c_{n}=c_{n-1}=\cdots=c_{3}=\frac{b_{3}+b_{1}}{b_{2}}=5$.
Since $b_{n}-5 b_{n-1}+b_{n-2}=0$, the characteristic equation is $x^{2}-5 x+1=0$, with roots $x_{1,2}=\frac{5 \pm \sqrt{21}}{2}$, so we set $b_{n}=\lambda_{1}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\lambda_{2}\left(\frac{5-\sqrt{21}}{2}\right)^{n}$.

Given $b_{1}=1, b_{2}=2$ and $b_{n}-5 b_{n-1}+b_{n-2}=0$, define $b_{0}=3$. Thus, $3=b_{0}=\lambda_{1}+\lambda_{2}$,
$$
\begin{array}{c}
1=b_{1}=\lambda_{1}\left(\frac{5+\sqrt{21}}{2}\right)+\lambda_{2}\left(\frac{5-\sqrt{21}}{2}\right) . \\
\text { Hence, } \lambda_{1}=\frac{63-13 \sqrt{21}}{42}, \lambda_{2}=\frac{63+13 \sqrt{21}}{42} . \\
\text { Therefore, } b_{n}=\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+ \\
\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}, \\
a_{n}=\frac{1}{n}\left[\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\right. \\
\left.\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}\right] .
\end{array}
$$