Series solution to Second-order ODE

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Homework Statement



Find the first four non-vanishing terms in a series solution of the form \sum from 0 to infinity of akxk for the initial value problem,

4xy''(x) + 6y'(x) + y(x) = 0, y(0) = 1 and y'(0) = -1/6

Homework Equations



The Attempt at a Solution



Taking the second derivative of the series solution form I obtained,

y = \sum from 0 to infinity of akxk
y' = \sum from 1 to infinity of kakxk-1
y'' = \sum from 0 to infinity of (k+2)(k+1)ak+2xk

Substituting into the ODE I obtained,

4\sum from 0 to infinity of (k+2)(k+1)ak+2xk+1 + 6\sum from 1 to infinity of kakxk-1 + \sum from 0 to infinity of akxk

Now, I am unsure of where to go from here. Does this become two separate series for \sum from 0 to infinity and \sum from 1 to infinity?
 
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No, it should be written as one series. Note that your series from 1 to infinity can just as well be written as from 0 to infinity, because kak is zero when k=0.
 
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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