14. The sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies: $a_{1}=1, a_{2}=1, a_{n+1}=\frac{n^{2} a_{n}^{2}+5}{\left(n^{2}-1\right) a_{n-1}}(n \geq 2)$, find the general term formula for $a_{n}$.
Solve the system given by $\left(n^{2}-1\right) a_{n-1} a_{n+1}=n^{2} a_{n}^{2}+5,\left((n-1)^{2}-1\right) a_{n-2} a_{n}=(n-1)^{2} a_{n-1}^{2}+5$, subtracting the two equations yields
$$
\left(n^{2}-1\right) a_{n-1} a_{n+1}-n(n-2) a_{n-2} a_{n}=n^{2} a_{n}^{2}-(n-1)^{2} a_{n-1}^{2},
$$

which simplifies to $(n-1) a_{n-1}\left((n+1) a_{n+1}+(n-1) a_{n-1}\right)=n a_{n}\left(n a_{n}+(n-2) a_{n-2}\right)$. Let $b_{n}=n a_{n}(n \geq 1)$, then we have $b_{n-1}\left(b_{n+1}+b_{n-1}\right)=b_{n}\left(b_{n}+b_{n-2}\right)$,
which simplifies to $\frac{b_{n+1}+b_{n-1}}{b_{n}}=\frac{b_{n}+b_{n-2}}{b_{n-1}}$. $\qquad$
Let $c_{n}=\frac{b_{n}+b_{n-2}}{b_{n-1}}(n \geq 3)$, given $a_{1}=1, a_{2}=1, a_{3}=3$, we get $b_{1}=1, b_{2}=2, b_{3}=9$,
thus $c_{n}=c_{n-1}=\cdots=c_{3}=\frac{b_{3}+b_{1}}{b_{2}}=5$. $\qquad$
Since $b_{n}-5 b_{n-1}+b_{n-2}=0$, the characteristic equation is $x^{2}-5 x+1=0$, with roots $x_{1,2}=\frac{5 \pm \sqrt{21}}{2}$, hence $b_{n}=\lambda_{1}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\lambda_{2}\left(\frac{5-\sqrt{21}}{2}\right)^{n}$. Given $b_{1}=1, b_{2}=2$ and $b_{n}-5 b_{n-1}+b_{n-2}=0$, define $b_{0}=3$,
thus $3=b_{0}=\lambda_{1}+\lambda_{2}, 1=b_{1}=\lambda_{1}\left(\frac{5+\sqrt{21}}{2}\right)+\lambda_{2}\left(\frac{5-\sqrt{21}}{2}\right)$,
which gives $\lambda_{1}=\frac{63-13 \sqrt{21}}{42}, \lambda_{2}=\frac{63+13 \sqrt{21}}{42}$, therefore
$$
\begin{array}{c}
b_{n}=\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}, \\
a_{n}=\frac{1}{n}\left(\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}\right) 。
\end{array}
$$
20 points