5. Given that $x$ and $y$ are unequal natural numbers, satisfying $x^{3}+19 y=y^{3}+19 x$.
Then the integer part of $\sqrt{x^{2}+y^{2}}$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1
5. B.

Assume $x>y$. From the given, we have $x^{3}-y^{3}=19(x-y)$.
Since $x$ and $y$ are both natural numbers, we have $x^{2}+x<x^{2}+xy+y^{2}=19<3x^{2}$. This leads to $x=3, y=2$.
Therefore, $\sqrt{x^{2}+y^{2}}=\sqrt{13}$, and its integer part is 3.