In a previous question, I was dealing with the following equation
$$x_1^2 \frac{d^2 A(x_1)}{dx_1^2}\frac{1}{A(x_1)} + x_1 \frac{d A(x_1)}{dx_1} \frac{1}{A(x_1)} + k^2 x_1^2 = -\frac{d^2 B(x_2)}{dx_2^2} \frac{1}{B(x_2)}$$
which can be written through the system
$$\begin{cases} \displaystyle x_1^2 \frac{d^2 A(x_1)}{dx_1^2}\frac{1}{A(x_1)} + x_1 \frac{d A(x_1)}{dx_1} \frac{1}{A(x_1)} + k^2 x_1^2 = C\\ \displaystyle -\frac{d^2 B(x_2)}{dx_2^2} \frac{1}{B(x_2)} = C \end{cases}$$
The general solution of the first equation is
$$F(x_1,x_2) = A(x_1)B(x_2) = \left[ K_1 J_C (k x_1) + K_2 Y_C (k x_2) \right] \left[ K_3 \sin (C x_2) + K_4 \cos (C x_2) \right]$$
where obviously $A(x_1)$ is the solution of the first equation of the system and $B(x_2)$ is the solution of the second equation of the system.
The initial equation (even if not arranged as so) is homogenous. I would say that just one value between $K_1$, $K_2$, $K_3$, $K_4$ is independent (the one that I would arbitrarily choose, as usual in homogeneous equations), but is it true? And if exist a relation between $K_1$, $K_2$, $K_3$, $K_4$, how can it be found?
When written with the constant $C$, the two equations aren't homogeneous, technically. Anyway, as explained in the accepted answer of the other post, one must have $C=0$.
This said, the two homogenous equations are of the second order so that their general solutions do involve two free constants, that can be determined from the initial conditions.
There is no reason to have a dependency between the constants of either solutions, and there are four degrees of freedom in total.