Choice of solution for Bessel's ODE when $\nu=0$

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Quick question: I was checking a solution for a PDE that went as follows. By variable separation, you find the ODE $$ \rho^2\,\xi''(\rho)+ \rho\,\xi'(\rho) + \kappa^2\rho^2\,\xi(\rho) = 0 $$ By inspection that is Bessel's equation for $\nu=0$. So far so good. Then, the solution proposes a solution in the form

\begin{equation}\label{eq:1} \xi(\rho) = A\,J_0(\kappa\rho) + B\,Y_0(\kappa\rho)\tag{1} \end{equation}

where $J_0$ is a Bessel Function of the first kind and $Y_0$ is one of the second kind. I read in Mathematical Methods for Physicists that

(...) a specific choice for a second solution is denoted $Y_ν(x)$ and is called a Bessel function of the second kind.

So, is equation \eqref{eq:1} so because an ODE of second order requires two solutions and those have to be $J_\nu$ and $Y_\nu$? Or is it because $\nu$ has to be zero and no other orthogonal polynomial were available but $Y_\nu$?

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Any combination

$$AJ_{0}\left(\kappa\rho\right) + BY_{0}\left(\kappa\rho\right)$$

with $B\neq0$ would be a second solution. Setting $A=0$ is a specific choice.