x-2-1-y-n-n-1-y-0-n-N-Find-solution-that-can-be-expanded-in-series-help-me-

Question Number 137129 by Chhing last updated on 30/Mar/21

(x^{2} - 1) y^{″} - n (n + 1) y = 0, n \in N

Find solution that can be expanded in series

help me

Answered by mathmax by abdo last updated on 31/Mar/21

y = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow y^{^{'}} = \sum_{n = 1}^{\infty} {na}_{n} x^{n - 1} and y^{^{″}} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

e \Rightarrow (x^{2} - 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - p (p + 1) \sum_{n = 0}^{\infty} a_{n} x^{n} = 0 \Rightarrow

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

\sum_{n = 2}^{\infty} {n (n - 1) - p (p + 1)} a_{n} x^{n} - p (p + 1) - p (p + 1) a_{1} x

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

{2 a}_{2} + {6 a}_{3} x + \sum_{n = 2}^{\infty} {(n (n - 1) - p (p + 1)) a_{n} - (n + 2) (n + 1) a_{n + 2}} x^{n}

- p (p + 1) - p (p + 1) a_{1} x + {2 a}_{2} + {6 a}_{3} x = 0 \Rightarrow

\Rightarrow (n^{2} - p^{2} - n - p) a_{n} - (n + 1) (n + 2) a_{n + 2} = 0 \forall n ⩾ 2 and

- p (p + 1) + {2 a}_{2} = 0 and ({6 a}_{3} - p (p + 1) a_{1}) = 0

\dots be continued \dots