Find the power series in x for the general solution of (1+2x^2)y"+7xy'+2y=0

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The discussion focuses on finding the power series solution for the differential equation (1+2x^2)y"+7xy'+2y=0. A correct recurrence relation a_{2n} = - \frac{2n+1}{n+1}a_n is identified, but the user struggles to derive the book's answer. The user explores methods for solving recurrences, including separating even and odd terms, but remains uncertain about the final form of the solution. Clarification is sought on how to derive the coefficients b_m and c_m from the established recurrence. The conversation highlights the complexity of solving differential equations using power series.
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Homework Statement


Find the power series in x for the general solution of (1+2x^2)y"+7xy'+2y=0.

Homework Equations


None.

The Attempt at a Solution


I'll post my whole work.
 
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This is my work, I have more to post.
 

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This work comes first, the above one comes after this one.
 

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But I don't know how to get to the book's answer.
 

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You have the correct recurrence relation a_{2n} = - \frac{2n+1}{n+1}a_n. You just haven't tried to solve it.

Recurrences of the form <br /> a_{n+1} = f(n)a_n have the solution <br /> a_n = a_0\prod_{k=0}^{n-1} f(k) where by convention <br /> \prod_{k=0}^{-1} f(k) = 1.
Recurrences of the form <br /> a_{n+2} = f(n)a_n can be turned into the above form by treating even and odd terms separately: First set n = 2m and b_m = a_{2m} to obtain <br /> b_{m+1} = f(2m)b_m and then set n = 2m+1 and c_m = a_{2m+1} to obtain <br /> c_{m+1} = f(2m+1)c_m.
 
But that's not the answer in the book. How do I get the answer in the book?
 
Are you saying that <br /> y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{m=0}^\infty b_mx^{2m} + \sum_{m=0}^\infty c_mx^{2m+1} with b_m and c_m obtained as I have suggested is not the answer in the book?

What do you get for b_m and c_m?
 

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