Ok so I am stuck. I need to get all the $n$'s to $=0$ but I can't reduce my series which has $n=2$ to $0$ because then I will have undone all my work in the first place to get all the $X^n$'s to the same power.. I am really confused on what to do.. Here is where I'm at now:
$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + \sum ^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - 6\sum ^{\infty}_{n=0} na_nX^n + 10 \sum ^{\infty}_{n=0}a_nX^n$$
any guidance would be great. Thx in advance
$$\sum ^{\infty}_{n=2} n(n-1)a_nX^n + (2a_2+6a_3+\sum ^{\infty}_{n=2} (n+2)(n+1)a_{n+2}X^n) - (a_1+6\sum ^{\infty}_{n=2} na_nX^n) + (a_1+10\sum ^{\infty}_{n=2}a_nX^n)$$
$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_nX^n + (n+2)(n+1)a_{n+2}X^n) - (6 na_nX^n) + (10a_nX^n)]$$
$$=(2a_2+6a_3)+\sum ^{\infty}_{n=2} [n(n-1)a_n + (n+2)(n+1)a_{n+2}) - (6 na_n) + (10a_n)]X^n$$
Now you go about solving your differential equation:
$$(2a_2+6a_3)=0\rightarrow a_2=-3a_3$$
$$\sum ^{\infty}_{n=2} [(n^2-7n+10))a_n + (n+2)(n+1)a_{n+2}) ]X^n=0\sum ^{\infty}_{n=2} [(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2}) ]X^n=0$$
$$(n-5)(n-2)a_n + (n+2)(n+1)a_{n+2})=0\rightarrow a_{n+2}=-\frac{(n-5)(n-2)}{(n+2)(n+1)}a_n$$