Please check Frobenius's method

[ognik]

Hi, thought I had this done, but I get a different answer from my book; I can see where we are different, so would appreciate if someone could tell me what I'm doing wrong here...

ODE is $ \d{}{x}[(1-x^2)\d{u}{x}]+\alpha u + \beta x^2u = 0 $, which I rewrote as $ (1-x^2)u'' -2xu' + \alpha u + \beta x^2 u = 0 $

Letting $ u=\sum_{\lambda=0}^{\infty}{a}_{\lambda}{x}^{k+\lambda} $ (any k)

I get the 3-term recursion relation below, using:
$ \lambda = j+2, $ for all $ k+ \lambda-2 $ power terms,
$ \lambda = j, $ for all $ k+ \lambda $ terms
$ \lambda = j-2, $ for all $ k+\lambda + 2 $ terms

$ {a}_{j+2}(k+j+2)(k+j+1)x^{k+j} - {a}_{j}(k+j)(k+j-1)x^{k+j} + \alpha{a}_{j} x^{k+j} + \beta {a}_{j-2}x^{k+j} = 0 $

But the book shows something different, for the 2nd term they have $ {a}_{j}(k+j)(k+j+1)x^{k+j} $ ?

I thought I must have a silly mistake, but after hours can't find it - hope someone can?
(PS: does the ODE have a name, like Legendre's etc?)

Found the silly mistake - would still like to know if the original ODE has a special name?

 

[jvicens]

Hello,

Thank you for posting your question on the forum. The original ODE does not have a specific name, but it is a second-order linear ordinary differential equation with variable coefficients. As for your mistake, it seems that you have a typo in your second term. It should be $ {a}_{j}(k+j)(k+j+1)x^{k+j} $ instead of $ {a}_{j}(k+j)(k+j-1)x^{k+j} $. This is because when you substitute $ \lambda = j $ in your general solution, you get $ {a}_{j}(k+j)(k+j+1)x^{k+j} $ for the $ k+\lambda $ term. I hope this helps clear up your confusion. Let me know if you have any other questions. Good luck with your work!