5. Let $\sqrt{x}+\sqrt{y}=\sqrt{2009}$, and $0<x<y$. Then the number of different integer pairs $(x, y)$ that satisfy this equation is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
5.C.

From $2009=49 \times 41$, we get $\sqrt{2009}=7 \sqrt{41}$.
Also, $0<x<y$, so $\sqrt{2009}$ can be rewritten as
$$
\begin{array}{l}
\sqrt{2009}=\sqrt{41}+6 \sqrt{41}=2 \sqrt{41}+5 \sqrt{41} \\
=3 \sqrt{41}+4 \sqrt{41},
\end{array}
$$

which means
$$
\begin{array}{l}
\sqrt{2009}=\sqrt{41}+\sqrt{1476} \\
=\sqrt{164}+\sqrt{1025}=\sqrt{369}+\sqrt{656} .
\end{array}
$$

Therefore, the integer pairs $(x, y)$ that satisfy the conditions are $(41,1476)$, $(164,1025)$, and $(369,656)$. There are 3 pairs in total.