2. Given $x \geqslant 0, x^{2}+(y-4)^{2} \leqslant 4$, let $u=$ $\frac{x^{2}+\sqrt{3} x y+2 y^{2}}{x^{2}+y^{2}}$. Then the range of $u$ is $\qquad$ .
$$
\begin{array}{l}
2.2 \leqslant w \leqslant \frac{5}{2} \\
w-\frac{1}{2}=\frac{\frac{1}{2}(x+\sqrt{3} y)^{2}}{x^{2}+y^{2}}=\frac{2\left(\frac{x+\sqrt{3} y}{2}\right)^{2}}{\left(\sqrt{x^{2}+y^{2}}\right)^{2}}
\end{array}
$$

As shown in Figure 4, let $P(x, y)$, the line $l: x+\sqrt{3} y=0$,
$$
PA \perp l \text{ at } A, \angle AOP =
$$
$\theta$, it is easy to know that $60^{\circ} \leqslant \theta \leqslant 90^{\circ}$.
$$
\begin{array}{l}
|PA|=\left|\frac{x+\sqrt{3} y}{2}\right|, \\
|PO|=\sqrt{x^{2}+y^{2}},
\end{array}
$$

Then $w-\frac{1}{2}=2 \sin ^{2} \theta \in\left[\frac{3}{2}, 2\right]$, i.e., $2 \leqslant w \leqslant \frac{5}{2}$.