2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$

Then the minimum value of $x+y$ is $\qquad$.
2. 2 .

Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$

When $x=\frac{1}{y}$,
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right)=1 .
$$

And $f(x)=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing, so $x \geqslant \frac{1}{y}$.
Thus, $x+y \geqslant y+\frac{1}{y} \geqslant 2$.
When $x=y=1$, the equality holds.