Solving ODE with Frobenius Form Solution

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The discussion focuses on solving an ordinary differential equation (ODE) using a Frobenius form solution, leading to a recurrence relation: a(k) = -2/(2k-1) a(k-1). The user seeks assistance in expressing this recurrence in terms of a(0) but is struggling to simplify it further. They note that the general form should include a (2n-1)! in the denominator and a (-2)^(n) in the numerator. There is a suggestion that a(0) might be zero, which could impact the overall solution. The user emphasizes the need for help in simplifying the expression rather than a complete explanation of the Frobenius method.
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I'm finding a solution to and ODE using a Frobenius form solution, I have worked through the question and have ended up with a recurrence of the form;

a(k) = -2/(2k-1) a(k-1)

I'm trying to find a general reccurence in terms of a(0) but am finding it a bit difficult, I know it has to have a (2n-1)! term on the bottom of the fraction and a (-2)^(n) term on top but I'm a bit stuck as to where to go after this.

Sorry I haven't Latex-ed anything, I don't know how!

Any help would be fantastic, thanks.
 
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good luck finding anyone who wants to actually write out frobenius method. If i remember correctly, there may be a point before the end where a(0) is established. and it could very well be 0. so heads up for that. otherwise, re-do it.
 
I'm not looking for anyone to write out Frobenius method for me, I just need some help in simplifying the expression in terms of a(0), so that I can get a general expression for a solution. The problem dosen't have anything to do with Frobenius method at all.
 

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