The general methodology for this involves assuming a solution of the form $$ y = \sum_{n=0}^\infty a_nx^{n+r}.$$
One normally keeps the index $0$ for the first and second derivatives. My question regards the case where $r=1$ in which case $$y' = \sum_{n=0}^\infty (n+1)a_nx^n $$ Which of the following is correct for $y''$?
$$y''= \sum_{n=0}^\infty n(n+1)a_nx^{n-1}$$ OR
$$y''= \sum_{n=1}^\infty n(n+1)a_nx^{n-1}$$
Classically the index would be incremented to $n=1$ but the indicial equation is only popping up in front of $a_0$ when I keep the index $0$. I haven't seen the existence theorem for this I'm wondering if maybe this is supposed to be considered a Laurent series with all the negative coefficients $0$
I am unsure why you have an $r$ in your equation, as forces $y(0)$ to be $0$ since there is no constant term. So I will explain how the Frobenius method works. When using the Frobenius method, we simply assume the solution is analytic, which means that we can express it as $$y=\sum_{n=0}^{\infty} a_nx^n$$ The constant term then drops out when we take the derivative term by term to get $$y' = \sum_{n=1}^{\infty}na_nx^{n-1} =\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$$ Then taking the derivative again, the constant term drops out again and we get $$y'' = \sum_{n=1}^{\infty} (n+1)(n)a_{n+1}x^{{n-1}} = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}$$ And in general $$y^{(m)}=\sum_{n=0}^{\infty}\bigg(\prod_{i=1}^{m}(n+i)\bigg) a_{n+m}x^n$$ So onto your question, your two sums are equivalent as the first term of your sum is simply $0$, as it was a constant which was differentiated. Hoepfully this helps!