Question-120631

Question Number 120631 by 77731 last updated on 01/Nov/20

Answered by mathmax by abdo last updated on 01/Nov/20

let determine a solution at formy = \sum_{n = 0}^{\infty} a_{n} x^{n}

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

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

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

{\begin{cases} 2 (n + 1) (n + 2) a_{n + 2} - a_{n - 1} + a_{n} = 0 \forall n ⩾ 1 \\ {4 a}_{2} + a_{0} = 0 \end{cases}

\Rightarrow {\begin{cases} a_{2} = - \\ 2 (n + 1) (n + 2) a_{n + 2} = a_{n - 1} - a_{n} \end{cases}

we have a_{n - 1} - a_{n} = 2 (n + 1) (n + 2) a_{n + 2} \Rightarrow

\sum_{k = 1}^{n} (a_{k - 1} - a_{k}) = 2 \sum_{k = 1}^{n} (k + 1) (k + 2) a_{k + 2} \Rightarrow

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

a_{n} = a_{0} - 2 \sum_{k = 1}^{n} (k + 1) (k + 2) a_{k + 2} \dots . . be contonued \dots