$$y_1''(x) - \frac{C_1}{x}y_1'(x)= C_2 y_1(x) - C_3 y_2(x)$$ $$y_2''(x) - \frac{C_4}{x}y_2'(x)= C_2 y_2(x) - C_3 y_1(x)$$
At the given boundary conditions:
$$ y_1(0) = y_2(0) = c $$ $$ y_1(\infty) = y_2 (\infty) = 0$$
I have tried to solve the problem as follow:
$$ y_1(x) = U + V $$ and $$ y_2(x) = U - V $$
and I got the following solution but it doesn't satisfy the problem, so I wonder if anyone can help.
$$ U = x^{\frac{1+C_1}{2}} * (k_1 * I_{\frac{1+C_1}{2}}(x\sqrt{C_2 - C_3}) + k_2 * K_{\frac{1+C_1}{2}}(x\sqrt{C_2 - C_3}))$$ $$ V = x^{\frac{1+C_2}{2}} * (k_3 * I_{\frac{1+C_2}{2}}(x\sqrt{C_2 + C_3}) + k_4 * K_{\frac{1+C_2}{2}}(x\sqrt{C_2 + C_3}))$$