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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 $$ \sum_{n=2}^\infty a_nn(n-1)x^{n-2} + x^3\sum_{n=2}^\infty a_nn(n-1)x^{n-2} -6x\sum_{n=0}^\infty a_nx^n$$ 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.

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  • $\begingroup$ It seems to be correct. My assumption is that the problem is formulated in a misleading way and you already solved it. $\endgroup$ – maxmilgram Jan 20 at 9:23
  • $\begingroup$ That being said, obviously you can create arbitrary many solutions by taking linear combinations of $y_1$ and $y_2$ but that seems a bit pointless. $\endgroup$ – maxmilgram Jan 20 at 9:25
  • $\begingroup$ $y=c_1 (1+x^3)$ is a solution of the equation. $\endgroup$ – Claude Leibovici Jan 20 at 9:27
  • $\begingroup$ I'd say your solution is correct as maxmilgram wrote, and the first three non-zero terms in a power-series expansion of $y$ are simply $y(x) \simeq a_0 + a_1 x + a_0 x^3$. $\endgroup$ – Christoph Jan 20 at 9:45
  • $\begingroup$ $y \left( x \right) =(y \left( 0 \right) +\mbox {D} \left( y \right) \left( 0 \right) x+y \left( 0 \right) {x}^{3}+{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{2}}{x}^{4}-{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{14}}{x}^{7}+{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{35}}{x}^{10}-{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{65}}{x}^{13}+O \left( {x}^{16} \right) ) $ $\endgroup$ – Aleksas Domarkas Jan 21 at 8:45
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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 $|x| < 1$

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