2. Given $x, y \in \mathbf{R}_{+}$, and
$$
x-3 \sqrt{x+1}=3 \sqrt{y+2}-y \text {. }
$$

then the maximum value of $x+y$ is $\qquad$
$$
2.9+3 \sqrt{15} \text {. }
$$

From $x+y=3(\sqrt{y+2}+\sqrt{x+1})$
$$
\leqslant 3 \sqrt{2(x+y+3)} \text {, }
$$

we get
$$
\begin{array}{l}
(x+y)^{2} \leqslant 18(x+y+3) \\
\Rightarrow(x+y)^{2}-18(x+y) \quad \leqslant 0 .
\end{array}
$$

Solving, we get $x+y \leqslant 9+3 \sqrt{15}$.
The maximum value $9+3 \sqrt{15}$ is achieved when $x=5+\frac{3 \sqrt{15}}{2}, y=4+\frac{3 \sqrt{15}}{2}$.