3. Let non-zero real numbers $a, b$ satisfy $a^{2}+b^{2}=25$. If the function $y=\frac{a x+b}{x^{2}+1}$ has a maximum value $y_{1}$ and a minimum value $y_{2}$, then $y_{1}-y_{2}=$ $\qquad$
3.5.

From
$$
\begin{aligned}
y & =\frac{a x+b}{x^{2}+1} \Rightarrow y x^{2}-a x+y-b=0 \\
& \Rightarrow \Delta=a^{2}-4 y(y-b) \geqslant 0 .
\end{aligned}
$$

Thus, $y_{2}$ and $y_{1}$ are the two roots of $a^{2}-4 y(y-b)=0$, at this point,
$$
\Delta_{1}=16 b^{2}+16 a^{2}=400>0 .
$$

Therefore, $y_{2}$ and $y_{1}$ always exist.
So $y_{1}-y_{2}=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}$
$$
=\sqrt{b^{2}-\left(b^{2}-25\right)}=5 \text {. }
$$