The following equation relates $A(x_1)$ (which is function of the variable $x_1$ only) and $B(x_2)$ (which is function of the variable $x_2$ only):
$$x_1^2 \frac{d^2 A(x_1)}{dx_1^2}\frac{1}{A(x_1)} + x_1 \frac{d A(x_1)}{dx_1} \frac{1}{A(x_1)} + k^2 x_1^2 = -\frac{d^2 B(x_2)}{dx_2^2} \frac{1}{B(x_2)}$$
where $x_1, x_2, k \in \mathbb{C}$. It must obviously hold for every couple $(x_1, x_2)$. Textbooks say that in such cases, when the LHS and the RHS are funcions of separate, independent variable, the only way to satisfy the equation is that each side is equal to the same constant. That is
$$\begin{cases} \displaystyle x_1^2 \frac{d^2 A(x_1)}{dx_1^2}\frac{1}{A(x_1)} + x_1 \frac{d A(x_1)}{dx_1} \frac{1}{A(x_1)} + k^2 x_1^2 = C\\ \displaystyle -\frac{d^2 B(x_2)}{dx_2^2} \frac{1}{B(x_2)} = C \end{cases}$$
1) Why is it so? Why is it necessary that the two sides are separately equal to a constant?
2) Are there any limitations on the value of $C$? For example, should it necessarily be an integer, or any complex number can be valid?
By the transitivity of the equality sign (to use a fancy term), we know that there exists some $F$ such that \begin{equation} x_1^2 \frac{d^2 A(x_1)}{dx_1^2}\frac{1}{A(x_1)} + x_1 \frac{d A(x_1)}{dx_1} \frac{1}{A(x_1)} + k^2 x_1^2 = F \qquad(1) \end{equation} and \begin{equation} F = -\frac{d^2 B(x_2)}{dx_2^2} \frac{1}{B(x_2)}. \qquad (2) \end{equation} If we take the partial derivative to $x_2$ of $(1)$, we obtain \begin{equation} 0 = \frac{\partial F}{\partial x_2}. \end{equation} Taking the partial derivative to $x_1$ of $(2)$, we likewise obtain \begin{equation} \frac{\partial F}{\partial x_1} = 0. \end{equation} We see that $F$ must be constant both in $x_1$ and $x_2$. Since those are the only two dynamical variables available, we conclude that $F$ is equal to a constant.
A priori, there are no restrictions on the value of $F$ (or $C$, as in your question). However, if in your case the value $x_1 = 0$ is allowed, from $(1)$ we can conclude that $F = 0$, since $(1)$ must hold for every value of $x_1$, so in particular for $x_1 = 0$ (assuming $A(x_1)$ doesn't have a pole at $x_1$, that is).