$\left(x^2+2\right)y''\:+\:xy'\:-\:y=0$
What's next after this?
$\sum _{n=2}^{\infty }\:n\left(n-1\right)a_nx^n+2\:\sum _{n=0}^{\infty \:}\left(n+2\right)\left(n-1\right)a_{n+2}x^n+\sum _{n=2}^{\infty \:\:}na_n^{\:}x^n\:-\:\sum _{n=0}^{\infty \:}a_n^{\:}x^n = 0.$
Restarting from scratch, we assume that the DE $$ (x^2+2)y''+xy'-y=0 \tag{1}$$ has an analytic solution in a neighbourhood of zero, given by $$ y(x) = \sum_{n\geq 0} a_n x^n \tag{2} $$ In terms of the coefficients, $(1)$ translates into:
$$ (x^2+2)\sum_{n\geq 2} n(n-1)a_n x^{n-2} + x\sum_{n\geq 1}na_n x^{n-1} - \sum_{n\geq 0}a_n x^n = 0 \tag{3}$$ that is equivalent to: $$ \sum_{n\geq 2}n(n-1)a_n x^n +\sum_{n\geq 0}2(n+2)(n+1)a_{n+2}x^n + \sum_{n\geq 1} n a_n x^n -\sum_{n\geq 0}a_n x^n = 0 \tag{4} $$ Now we consider what is the coefficient of $x^n$ in the last expression and we get that,
for every $n\geq 2$, $$ n(n-1) a_n + 2(n+2)(n+1) a_{n+2} + (n-1)a_n = 0\tag{5} $$ hence: $$ \frac{a_{n+2}}{a_n} = - \frac{(n-1)}{2(n+2)} \tag{6} $$ and by induction: $$ a_{n+2k} = \frac{(-1)^k}{2^k}\cdot\frac{(n+2k-3)(n+2k-5)\cdot\ldots\cdot(n-1)}{(n+2k)(n+2k-2)\cdot\ldots\cdot(n+2)}\,a_n \tag{7}$$ so a solution of $(1)$ is given by a hypergeometric function.