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[tex](3x^4+4x^2+1)y'' + (6x^3-2x)y' -(6x^2-2)y=0[/tex]

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

[tex]y=a_0(1-x^2+x^4-x^6+...)+a_1x[/tex]

from the recurrence relation:

[tex]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}[/tex]

Now it is clear that the two independent homogeoneous solutions are

[tex]y_1=1-x^2+x^4-x^6+... = \frac{1}{1+x^2}[/tex]

[tex]y_2=x[/tex]

and the radius of convergence of the series

[tex]y_1=1-x^2+x^4-x^6+... [/tex]

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)?