Find power series representations of the general solution

Shackleford
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Homework Statement



(1+x2) y'' + 2xy' = 0 in powers of x

Homework Equations



y'' = \sum_{n=2}^{\infty} (n-1)na_nx^{n-2}

y' = \sum_{n=1}^{\infty} na_nx^{n-1}

The Attempt at a Solution



(1+x2) y'' + 2xy' =

(1+x^2) \sum_{n=2}^{\infty} (n-1)na_nx^{n-2} + 2x \sum_{n=1}^{\infty} na_nx^{n-1} = 0

(1+x^2) \sum_{n=2}^{\infty} (n-1)na_nx^{n-2} + 2x \sum_{n=2}^{\infty} (n-1)na_nx^{n} + 2<br /> \sum_{n=1}^{\infty} na_nx^{n} = 0

2a_2 + 6a_3x + 2a_1x + \sum_{n=2}^{\infty} [(m+1)(m+2)a_mx + (m-1)ma_m + 2ma_m] x^{m}

a_0 = a_0 \\<br /> a_1 = a_1 \\<br /> 6a_3 + 2a_1 = 0 \\<br /> 12a_4 + 6a_2 = 0, a_4 = 0 \\ <br /> 20a_5 + 12a_3 = 0 \\<br />

a_{2n} = 0, a_{2n+1} = (-1)^n\frac{a_1}{2n+1}
 
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Hi @Shackleford . We're all waiting for the other shoe to drop ... :wink:

... is there a question coming?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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