<s>[INST] Not possible to find non-zero terms of series expansion? I've been asked to compute the first 3 nonzero terms of a power series expansion about x=0 for two linearly independent solutions to the ODE:
$$(1+x^3)y''- 6xy =0 $$
I have tried to solve this many different ways and continue to get the same solution, which does not allow me to find three nonzero terms from two linearly independent solutions
I have used $$y(x) = \sum_{n=0}^\infty a_nx^n $$
$$ y'(x)=\sum_{n=1}^\infty a_nnx^{n-1} $$
$$ y''(x)=\sum_{n=2}^\infty a_nn(n-1)x^{n-2} $$
to obtain the series form 
$$ \sum_{n=2}^\infty a_nn(n-1)x^{n-2} + x^3\sum_{n=2}^\infty a_nn(n-1)x^{n-2} -6x\sum_{n=0}^\infty a_nx^n$$
Add in x terms: 
$$ \sum_{n=2}^\infty a_nn(n-1)x^{n-2} + \sum_{n=2}^\infty a_nn(n-1)x^{n+1} -\sum_{n=0}^\infty 6a_nx^{n+1}$$
Set all the powers of x equal to n:
$$ \sum_{n=0}^\infty a_{n+2}(n+2)(n+1)x^{n} + \sum_{n=3}^\infty a_{n-1}(n-1)(n-2)x^{n} -\sum_{n=1}^\infty 6a_{n-1}x^{n}$$
Peel off terms to have all series start at n=3:
$$ 2a_2x^0+6a_3x^1-6a_0x^1+12a_4x^2-6a_1x^2$$ $$+$$
$$\sum_{n=3}^\infty [a_{n+2}(n+2)(n+1)+a_{n-1}(n-1)(n-2)-6a_{n-1}]x^n=0$$
From this I have deduced:
$ x^0 : a_2=0 $
$ x^1 : a_3=a_0 $
$ x^2 : a_4=\frac{a_1}{2} $
$ x^n, n\geq3 : $
\begin{align}
 & a_{n+2}=-\frac{a_{n-1}(n^2-3n+2-6)}{(n+2)(n+1)}\\
 & = -\frac{a_{n-1}(n+1)(n-4)}{(n+1)(n+2)}\\
 & = -\frac{a_{n-1}(n-4)}{n+2}\\
\end{align}
Using the recursion equation above I obtained these terms:
$n=3 : $
$$ a_5=\frac{a_2}{5}=0 $$
$n=4 : $
$$ a_6=a_3(4-4)=0 $$
$n=5 : $
$$ a_7=\frac{-a_4}{7}=\frac{-a_1}{14} $$
$n=6 : $
$$ a_8=\frac{-2a_5}{5}=\frac{-a_2}{20}=0$$
$n=7 : $
$$ a_9=\frac{-3a_6}{9}=0$$
$n=8 : $
$$ a_10=\frac{-4a_7}{12}=\frac{a_4}{21}=\frac{a_1}{42} $$
$n=9 : $
$$ a_11=\frac{-5a_8}{11}=\frac{a_2}{44}=0 $$
$n=10 : $
$$ a_12=\frac{-6a_9}{12}=\frac{-a_2}{2}=0 $$
$n=11 : $
$$ a_13=\frac{-7a_10}{13}=\frac{-7a_4}{273}=\frac{-a_1}{78} $$
So, every second and third term equal zer0. 
My solution is: 
$$ y_1(x)=a_0(1+x^3) $$
$$ y_2(x)=a_1\big(x+\frac{x^4}{2}-\frac{x^7}{14}+\frac{x^{10}}{42} - ...) $$
Can someone please tell me what I am doing wrong here?
I cannot come up with three non zero terms from $y_1$ as there are only two terms, and everything else is zero. 
When I try to apply the ratio test to this series solution I get inconclusive results as well which further makes me think my solution is incorrect...
Any help or advice would be very greatly appreciated. 
 [/INST] To verify the correctness of your solution I made the following MATHEMATICA script with the results shown below.
Operator = (1 + x^3) D[#, {x, 2}] - 6 x # &;
n = 20;
Sumb = Sum[Subscript[alpha, k] x^k, {k, 0, n}];
res = Operator[Sumb] /. {Subscript[alpha, 0] -> Subscript[a, 0], 
Subscript[alpha, 1] -> Subscript[a, 1]};
coefs = CoefficientList[res, x];
equs = Thread[coefs == 0];

For[k = 1; Alphas = {}; equsk = equs[[1]]; subs = {}, 
    k <= Length[equs] - 1, k++, 
    solalphak = Solve[equsk, Subscript[alpha, k + 1]];
    AppendTo[Alphas, Subscript[alpha, k + 1] /. solalphak][[1]];
    AppendTo[subs, solalphak];
    equsk = equs[[k + 1]] /. Flatten[subs]
]

Alphas0 = Flatten[Alphas];
series = Subscript[a, 0] + x Subscript[a, 1] + 
Sum[Alphas0[[k]] x^(k + 1), {k, 1, Length[Alphas0] - 2}]

$$
y(x) = a_0(1+x^3)+a_1\left(x+\frac{x^4}{2}-\frac{x^7}{14}+\frac{x^{10}}{35}-\frac{x^{13}}{65}+\frac{x^{16}}{104}-\frac{x^{19}}{152}+\cdots +\right)
$$
This is a convergent series for $|x| < 1$
</s>