According to this site, if $v$ is non-integer the solution of
$$x^2y''+xy'+(x^2-v^2)y=0\tag1$$
is
$$y(x)=C_1J_v(x)+C_2J_{-v}(x).\tag2$$
Where $J_v(x)$ is the Bessel function of the first kind.
In my BETA mathematics handbook, they have this form of the Bessel differential equation:
$$x^2y''+xy'+(a^2x^2-v^2)y=0,\tag3$$
and they claim the solution is
$$y(x)=C_1J_v(ax)+C_2Y_v(ax),\tag4$$
Where $Y_v(x)$ is the Weber function. Setting $a=1$ then$(1)=(3)$ but still the solution number $(4)$ will be different from solution $(2)$:
$$y(x)=C_1J_v(x)+C_2Y_v(x).$$
Does this imply that $J_{-v}=Y_v?$
For non-integer $v$ the following relation holds:
$$ Y_v(x)=\frac{J_v(x)\cos v\pi-J_{-v}(x)}{\sin v\pi} $$
so that both answers are correct up to redefinition of constants $C_1$ and $C_2$.
The advantage of the equation $(4)$ is the fact that it is valid for integer $v$ as well.