Convergence of Series Solution

To solve the 2nd order ode:

$$(3x^4+4x^2+1)y'' + (6x^3-2x)y' -(6x^2-2)y=0$$

I used a Taylor series expansion around x=0, and I got the general solution:

$$y=a_0(1-x^2+x^4-x^6+...)+a_1x$$

from the recurrence relation:

$$a_{n+2}=-\left[\frac{4n(n-1)-2n+2}{(n+2)(n+1)}\right]a_n-\left[\frac{3(n-2)(n-3)+6(n-2)-6}{(n+2)(n+1)}\right]a_{n-2}$$

Now it is clear that the two independent homogeoneous solutions are

$$y_1=1-x^2+x^4-x^6+... = \frac{1}{1+x^2}$$
$$y_2=x$$

and the radius of convergence of the series

$$y_1=1-x^2+x^4-x^6+...$$

is 1,

however,

using the recurrence relation I'm not being able to show that the radius of convergence of the series is indeed 1.

Can anyone help me to show this from the recurrence relation I obtained (I'm trying to use the ratio test, but I get another radius of convergence rather than 1)?

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In the theory of DE where should we compute the radius of convergence. Is it from

(a) the differential equation itself
(b) from the recurrence relation or from
(c) the series solutions ?

And do we expect the answers from (a) , (b) , and (c) equal ?

In the theory of DE where should we compute the radius of convergence. Is it from

(a) the differential equation itself
(b) from the recurrence relation or from
(c) the series solutions ?

And do we expect the answers from (a) , (b) , and (c) equal ?
O: I was always taught that it was from the recurrence relation . . . the recurrence relation is what defines the solution series itself, so wouldn't it be the same thing? And I've never considered computing the radius of convergence from the differential equation itself, unless its considering which points of the equation are regular and which are singular, and assuming that a series solution will converge from the point it's centered around until the nearest singular point of the equation . . . though I don't know why that's true, if it is true :P

I have no idea where should we compute the radius of convergence and I would also like to know about it. You could be right trying to compute it from the recurrence relation but it looks quite difficult to do it.

My opinion is that we compute the radius of convergence from the solution obtained.

What about from the DE itself ? If we write the given DE as
$$y\prime{} \prime{} + \frac{6x^3-2x}{3x^4+4x^2+1}y\prime{} -\frac{6x^2-2}{3x^4+4x^2+1}y=0$$

Since $$3x^4+4x^2+1 \equiv (1+x^2)(1+3x^2)$$ so the coefficients of the DE only converges for
$$-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$$

Or am I talking nonsense here :yuck:

What about from the DE itself ? If we write the given DE as
$$y\prime{} \prime{} + \frac{6x^3-2x}{3x^4+4x^2+1}y\prime{} -\frac{6x^2-2}{3x^4+4x^2+1}y=0$$

Since $$3x^4+4x^2+1 \equiv (1+x^2)(1+3x^2)$$ so the coefficients of the DE only converges for
$$-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$$

Or am I talking nonsense here :yuck:
The theorem says that the radius of convergence for the solution is at least the minimum radius of convergence of the coefficients. It can happen that the solution has larger radius of convergence. In this case, the solution $$x$$ even has infinite radius of convergence.

The theorem says that the radius of convergence for the solution is at least the minimum radius of convergence of the coefficients. It can happen that the solution has larger radius of convergence. In this case, the solution $$x$$ even has infinite radius of convergence.
That shed some light on the problem. So can I conclude that

i) we determine the radius of convergence from the final solution.

ii) the radius of convergence for the following DE are the same.

$$(3x^4+4x^2+1)y\prime{} \prime{} + (6x^3-2x)y' -(6x^2-2)y=0$$

$$y\prime{} \prime{} + \frac{6x^3-2x}{3x^4+4x^2+1}y\prime{} -\frac{6x^2-2}{3x^4+4x^2+1}y=0$$