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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
$$ sum_{n=2}^infty a_nn(n-1)x^{n-2} + x^3sum_{n=2}^infty a_nn(n-1)x^{n-2} -6xsum_{n=0}^infty a_nx^n$$
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, ngeq3 : $

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:
$$ y_1(x)=a_0(1+x^3) $$
$$ y_2(x)=a_1big(x+frac{x^4}{2}-frac{x^7}{14}+frac{x^{10}}{42} - ...) $$

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.

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}]

$$
y(x) = a_0(1+x^3)+a_1left(x+frac{x^4}{2}-frac{x^7}{14}+frac{x^{10}}{35}-frac{x^{13}}{65}+frac{x^{16}}{104}-frac{x^{19}}{152}+cdots +right)
$$

This is a convergent series for $|x| < 1$