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We have the following second order differential equation:

$$(x^2 - 3)y'' + 2xy' = 0$$

I want to solve it by the power series method.

I am assuming that the solution can be expanded in a power series around zero:

$$y(x) = \sum_{n=0}^{\infty} a_n x^n$$

After differentiation and plugging into the differential equation I get the recurrence relation (I have checked it and it is OK):

$$a_{n+2} = \frac{n(n-1) + 2n}{3(n+2)(n+1)}a_{n}$$

Note how each coefficient is related to the second one before it. This means we'll get (separately) both even and odd coefficients.

Let's draw our attention to the odd case in this question.

For the odd case we get that $n= 2K + 1$. Then our recurrence formula for the odd case is:

$$a_{2k+3} = \frac{(2K+1)}{3(2K+3)}a_{2k+1}$$

Here comes the issue: finding the power series for the odd solution. We expect this power series to have the following form:

$$y(x) = \sum_{K=0}^{\infty} a_{2K+1} x^{2K+1}$$

The method I use is the following:

1) Rewrite the coefficient on the RHS of the equation ($a_{2K+1}$ in this case) in the same form that the coefficient on the LHS.

Well, to do so we gotta do the following:

$$a_{2k+1} = \frac{((2K+1)-2)}{(3(2K+3)-2)}a_{2k+1-2} = \frac{(2K-1)}{3(2K+1)}a_{2k-1}$$

2) Keep applying 1) until you see the pattern.

$$a_{2k+1} = \frac{(2K-1)}{3(2K+1)}a_{2k-1} = \frac{(2K-1)}{3(2K+1)}\frac{(2K-3)}{3(2K-1)}a_{2k-3}=... $$

Thus:

$$a_{2k+1} = \frac{(2K+1)!!}{3^K(2K+3)!!}a_{1} = \frac{1}{3^K(2K+3)}a_{1}$$

Therefore I get the series:

$$y(x) = \sum_{K=1}^{\infty} \frac{a_{1}}{3^K(2K+3)} x^{2K+1}$$

But the solution states that the series is:

$$y(x) = \sum_{K=1}^{\infty} \frac{a_{1}}{3^K(2K+1)} x^{2K+1}$$

Where did I get wrong?

PS: Note that I want to avoid methods using tables, getting all the coefficients and guessing by them the power series, as I have been told that with more complicated power series that is not the best way to approach the problem. Please let me know if there's another useful method if needed be.

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    $\begingroup$ You have the right work, but your double factorials are wrong. How could the bottom be $(2K+3)!!$ when the largest term is $(2K+1)$ ? $\endgroup$ Oct 28 '19 at 18:24
  • $\begingroup$ Absolutely! It is: $a_{2k+1} = \frac{(2K-1)!!}{3^K(2K+1)!!}a_{1} = \frac{1}{3^K(2K+1)}a_{1}$ Thanks for spotting it. $\endgroup$
    – JD_PM
    Oct 28 '19 at 18:39
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I would start by reducing the given ODE to a first order DE:

\begin{align*} & (x^{2}-3)y^{\prime\prime} + 2xy^{\prime} = 0 \Longleftrightarrow (x^{2}y^{\prime\prime} + 2xy^{\prime}) - 3y^{\prime\prime} = 0 \Longleftrightarrow\\\\ & (x^{2}y^{\prime})^{\prime} - 3y^{\prime\prime} = 0 \Longleftrightarrow x^{2}y^{\prime} - 3y^{\prime} = k \Longleftrightarrow y^{\prime} (x^{2}-3) = k \end{align*}

which is easier to cope with.

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