# 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.

• It seems to be correct. My assumption is that the problem is formulated in a misleading way and you already solved it. – maxmilgram Jan 20 at 9:23
• 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. – maxmilgram Jan 20 at 9:25
• $y=c_1 (1+x^3)$ is a solution of the equation. – Claude Leibovici Jan 20 at 9:27
• 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 $y(x) \simeq a_0 + a_1 x + a_0 x^3$. – Christoph Jan 20 at 9:45
• $y \left( x \right) =(y \left( 0 \right) +\mbox {D} \left( y \right) \left( 0 \right) x+y \left( 0 \right) {x}^{3}+{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{2}}{x}^{4}-{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{14}}{x}^{7}+{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{35}}{x}^{10}-{\frac {\mbox {D} \left( y \right) \left( 0 \right) }{65}}{x}^{13}+O \left( {x}^{16} \right) )$ – Aleksas Domarkas Jan 21 at 8:45

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];

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 $$|x| < 1$$