Example 1 䒴 $\frac{4 x}{x^{2}-4}=\frac{a}{x+2}+\frac{b}{x-2}$, then the value of $a^{2}+$ $b^{2}$ is $\qquad$
(1996, Hope Cup National Mathematics Invitational Competition)
Solution: From the problem, we know that the given equation is not an equation in $x$, but an identity in $x$. From
$$
\frac{a}{x+2}+\frac{b}{x-2}=\frac{(a+b) x-2(a-b)}{x^{2}-4}
$$

we get the identity
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$

According to the conditions for the identity of polynomials, we get the system of equations about $a$ and $b$
$$
\left\{\begin{array}{l}
a+b=4, \\
a-b=0 .
\end{array}\right.
$$

Solving, we get $a=b=2$.
Therefore, $a^{2}+b^{2}=8$.