I have to find a Green function that obeys the following equation:
$$\frac{d^2G}{dx^2}+\frac{2}{x}\frac{dG}{dx}-\frac{l(l+1)}{x^2}G=\delta(x-a) \; (I)$$
With, $a>0$, $l$ is a integer, $0<x< \infty$, and the conditions $G(0;\xi)=G(\infty;\xi)=0$
It's easy to see that (I) is an Euler-Cauchy equation, and its homogeneous solution is like:
$$G= AX^{m_0} +BX^{m_1}$$ But I stopped there because some doubts arose.
Applying the conditions I get both equations that represent the Green equation ($G_{left}(x;\xi) \, \text{for} \; 0<\xi<x \; \text{and} \; G_{right}(x;\xi) \, \text{for} \; x<\xi<\infty) $ ? Or do I need to do something else?
Regarding the Dirac delta, i need to consider it to find the particular ODE solution? I thought of integrating (I) across the domain, but it seems to be unnecessary. Or use the Fourier transform, but that's too overkill.
Obs:I'm studying on my own.
The idea is that the Green's function will be of the form $$G(x) = f(x)H(x-a)$$ where $H$ is the Heaviside function and $f$ is the homogeneous solution that satisfies $f(a) = 0$ and $f'(a) = 1$ because $$G'(x) = f'(x)H(x-a) + f(a)\delta(x-a) = f'(x)H(x-a)$$ $$G''(x) = f''(x)H(x-a) + f'(a)\delta(x-a) = f''(x)H(x-a) + \delta(x-a)$$ which means that $$D(f(x)H(x-a)) = H(x-a)\cdot D(f(x)) + \delta(x-a) = \delta(x-a)$$ where $D$ is the differential operator that represents the DE.