let give the d.e. (1+x^2 )y^(′′) +3xy^′ +y =0find a solution y(x) deveppable at integr serie with∣x∣<1 .


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Question Number 34311 by prof Abdo imad last updated on 03/May/18

$l e t g i v e t h e d . e . (1 + x^{2}) y^{^{″}} + 3 {x y}^{^{'}} + y = 0 f i n d$ $a s o l u t i o n y (x) d e v e p p a b l e a t i n t e g r s e r i e$ $w i t h ∣ x ∣< 1 .$
	Answered by candre last updated on 04/May/18

	$y = \underset{i = 0}{\sum^{\infty}} a_{i} x^{i}$ $y^{'} = \underset{i = 0}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i}$ $y^{″} = \underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i}$ $(1 + x^{2}) \underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} + 3 x \underset{i = 0}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i} + \underset{i = 0}{\sum^{\infty}} a_{i} x^{i} = 0$ $\underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} + \underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i + 2} + \underset{i = 0}{\sum^{\infty}} 3 a_{i + 1} (i + 1) x^{i + 1} + \underset{i = 0}{\sum^{\infty}} a_{i} x^{i} = 0$ $\underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} + \underset{i = 2}{\sum^{\infty}} a_{i} i (i - 1) x^{i} + \underset{i = 1}{\sum^{\infty}} 3 a_{i} {i x}^{i} + \underset{i = 0}{\sum^{\infty}} a_{i} x^{i} = 0$ $2 a_{2} + a_{0} = 0$ $6 a_{3} + 4 a_{1} = 0$ $a_{i + 2} (i^{2} + 3 i + 2) + a_{i} (i^{2} + 2 i + 1) = 0 (i ⩾ 2)$
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