14) Given the sequence $a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies: $a_{1}=1, a_{2}=1, a_{n+1}=$ $\frac{n^{2} a_{n}^{2}+5}{\left(n^{2}-1\right) a_{n-1}}(n \geqslant 2)$, find the general term formula for $a_{n}$.
14 From the given, we have
$$
\begin{array}{c}
\left(n^{2}-1\right) a_{n-1} a_{n+1}=n^{2} a_{n}^{2}+5, \\
\left((n-1)^{2}-1\right) a_{n-2} a_{n}=(n-1)^{2} a_{n-1}^{2}+5,
\end{array}
$$

Subtracting the two equations, we get
$$
\left(n^{2}-1\right) a_{n-1} a_{n+1}-n(n-2) a_{n-2} a_{n}=n^{2} a_{n}^{2}-(n-1)^{2} a_{n-1}^{2},
$$

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 $b_{n}=n a_{n}(n \geqslant 1)$, then we have
$$
\begin{aligned}
b_{n-1}\left(b_{n+1}+b_{n-1}\right) & =b_{n}\left(b_{n}+b_{n-2}\right), \\
\frac{b_{n+1}+b_{n-1}}{b_{n}} & =\frac{b_{n}+b_{n-2}}{b_{n-1}} .
\end{aligned}
$$

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 get $b_{1}=1$, $b_{2}=2, b_{3}=9$, thus
$$
c_{n}=c_{n-1}=\cdots=c_{3}=\frac{b_{3}+b_{1}}{b_{2}}=5 .
$$

Since $b_{n}-5 b_{n-1}+b_{n-2}=0$, the characteristic equation is $x^{2}-5 x+1=0$, with roots $x_{1,2}=\frac{5 \pm \sqrt{21}}{2}$. Therefore, we can 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 $b_{0}=3$, then we have
$$
\begin{array}{l}
3=b_{0}=\lambda_{1}+\lambda_{2}, \\
1=b_{1}=\lambda_{1}\left(\frac{5+\sqrt{21}}{2}\right)+\lambda_{2}\left(\frac{5-\sqrt{21}}{2}\right),
\end{array}
$$

which gives
$$
\lambda_{1}=\frac{63-13 \sqrt{21}}{42}, \lambda_{2}=\frac{63+13 \sqrt{21}}{42},
$$

Therefore, we have
$$
\begin{array}{c}
b_{n}=\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}, \\
a_{n}=\frac{1}{n}\left(\frac{63-13 \sqrt{21}}{42}\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\frac{63+13 \sqrt{21}}{42}\left(\frac{5-\sqrt{21}}{2}\right)^{n}\right) .
\end{array}
$$