I'm trying to solve the following differential equation using Frobenius' method: 4xy''+2y''+y=0
I checked x=0 is a regular singular point and performed the substitution $y=x^{\lambda}\sum_{n=0}^{\infty}a_{n}x^n$
Then after multiplying everything through I got two series, one multiplying with x^n-1 and the other with x^n, if I shift the index of the latter n->n-1 and take the term corresponding to n=0 out of the first one I can put everything inside a sum that starts at n=1 and multiplies with x^n-1:
$\frac{a_{0}}{x}(4\lambda(\lambda -1)+2\lambda )) + \sum_{n=1}^{\infty }(a_{n}(4(n+\lambda)(n+\lambda-1)+(2n+2\lambda))+a_{n-1})x^{n-1}=0$
from the indicial equation I find indices 1/2 and 0. Then, using the first index, I get the recurrence relation $a_{n}=\frac{-a_{n-1}}{2n(2n+1)}$ which I believe is correct and gives me one of the solutions y1=Acos(x^(1/2))
However, if I had instead put everything inside a sum times x^n, by taking the n=0 term of the series out and then shifting the index with n->n+1 I get the following
$\frac{a_{0}}{x}(4\lambda(\lambda -1)+2\lambda )) + \sum_{n=0}^{\infty }(a_{n+1}(4(n+\lambda+1)(n+\lambda)+2(n+\lambda+1))+a_{n})x^{n}=0$
which when making each coefficient of x vanish gives me a different relation $a_{n+1}=\frac{-a_{n}}{4n^2+10n+6}$
I know I must be doing something wrong since they should both give me the same answer but shouldn't the two series be equivalent and give the same recurrence relations?
The relations are in fact one and the same, only written for different indices. If you take the first one and replace $\,n \mapsto \color{blue}{n+1}\,$ you get precisely the second one:
$$\require{cancel} a_{\color{blue}{n+1}}=\frac{-a_{\color{blue}{n+\bcancel{1}}-\bcancel{1}}}{2(\color{blue}{n+1})\big(2(\color{blue}{n+1})+1\big)} = \frac{-a_n}{(2n+2)(2n+3)} = \frac{-a_n}{4n^2+10n+6} $$