For:
First ODE:
$$ y_1''(x) + \frac{C_1}{x}y_1'(x)= C_2 y_1(x) + C_3 y_2(x)$$
Second ODE:
$$ y_2''(x) + \frac{C_4}{x}y_2'(x)= C_5 y_2(x) + C_6 y_1(x)$$
proceed as follows:
multiply the first by $C_{6}$ and the second by $C_{2}$ and subtract to obtain
$$C_{6} \, \left(y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} \right) = \lambda = C_{2} \left( y_{2}^{''} + \frac{C_{4}}{x} \, y_{2}^{'} \right) + (C_{6} C_{3} - C_{2} C_{5}) y_{2},$$
where $\lambda$ is a separation constant. Now,
\begin{align}
y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} &= \frac{\lambda}{C_{6}} \\
y_{2}^{''} + \frac{C_{1}}{x} \, y_{2}^{'} + \frac{C_{6}C_{3} - C_{2}C_{5}}{C_{2}} \, y_{2} &= \frac{\lambda}{C_{2}}.
\end{align}
The $y_{1}$ equation can be solved as follows:
\begin{align}
y_{1}^{''} + \frac{C_{1}}{x} \, y_{1}^{'} &= \frac{\lambda}{C_{6}} \\
\frac{1}{x^{C_{1}}} \, \frac{d}{dx} \left( x^{C_{1}} \, y_{1}^{'} \right) &= \frac{\lambda}{C_{6}} \\
x^{C_{1}} \, \frac{d y_{1}}{dx} &= \frac{\lambda \, x^{C_{1} + 1}}{C_{6} \, (C_{1} + 1)} + d_{0} \\
\frac{d y_{1}}{dx} &= \frac{\lambda \, x}{C_{6} \, (C_{1} + 1)} + \frac{d_{0}}{x^{C_{1}}}\\
y_{1} &= \frac{\lambda \, x^{2}}{2 \, C_{6} \, (C_{1} + 1)} - \frac{d_{0}}{(C_{1} -1) \, x^{C_{1}-1}} + d_{1}.
\end{align}
The second equation is solvable in terms of Bessel functions as seen by
$$f'' + \frac{a}{x} \, f' + b f = c$$
has the solution
$$f(x) = x^{(1-a)/2} \, \left(A_{0} \, J_{\beta}(\sqrt{b} \, x) + B_{0} \, Y_{\beta}(\sqrt{b} \, x) \right) + \frac{c}{b},$$
where $\beta = (1-a)/2$.
Given the conditions for the equations and the solution then the forms of solutions for $y_{1}$ and $y_{2}$ could be reduced.