(x^2 +2)y′′ + xy = 0


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Question Number 79761 by Najaah last updated on 28/Jan/20

$(x^{2} + 2) y^{″} + x y = 0$
Commented by abdomathmax last updated on 28/Jan/20
$solution developpable at?integr serie y(x)=Σ_(n=0) ^∞ a_n x^n ⇒y^′ =Σ_(n=1) ^∞ na_n x^(n−1) ⇒ y^((2)) (x)=Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) (e) ⇒(x^2 +2)Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) +Σ_(n=0) ^∞ a_n x^(n+1) =0 ⇒Σ_(n=2) ^∞ n(n−1)a_n x^n +2Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) +Σ_(n=1) ^∞ a_(n−1) x^n =0 ⇒ a_0 +Σ_(n=2) ^∞ {n(n−1)a_n +a_(n−1) }x^n +2a_2 +2Σ_(n=1) ^∞ (n+2)(n+1)a_(n+2) x^n =0 ⇒a_0 +2a_2 +Σ_(n=2) ^∞ {n(n−1)a_n +a_(n−1) +(n+2)(n+1)a_(n+2) }x^n =0 ⇒a_0 +2a_2 =0 and (n+2)(n+1)a_(n+2) +n(n−1)a_n +a_(n−1) =0 ∀n≥2 ...be continued...$
$s o l u t i o n d e v e l o p p a b l e a t ? i n t e g r s e r i e$ $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow y^{^{'}} = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1} \Rightarrow$ $y^{(2)} (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ $(e) \Rightarrow (x^{2} + 2) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 0$ $\Rightarrow \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} + 2 \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ $+ \sum_{n = 1}^{\infty} a_{n - 1} x^{n} = 0 \Rightarrow$ $a_{0} + \sum_{n = 2}^{\infty} {n (n - 1) a_{n} + a_{n - 1}} x^{n} + 2 a_{2}$ $+ 2 \sum_{n = 1}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} = 0$ $\Rightarrow a_{0} + 2 a_{2} + \sum_{n = 2}^{\infty} {n (n - 1) a_{n} + a_{n - 1} + (n + 2) (n + 1) a_{n + 2}} x^{n} = 0$ $\Rightarrow a_{0} + 2 a_{2} = 0 a n d$ $(n + 2) (n + 1) a_{n + 2} + n (n - 1) a_{n} + a_{n - 1} = 0 \forall n ⩾ 2$ $. . . b e c o n t i n u e d . . .$
	Answered by Henri Boucatchou last updated on 28/Jan/20

	$w e h a v e \frac{y^{″}}{y} = - \frac{x}{x^{2} + 2} \Rightarrow {l n y}^{'} = - \frac{1}{2} l n (x^{2} + 2)$ $\Rightarrow y^{'} = {(x^{2} + 2)}^{- 1 / 2}$ $\Rightarrow y = \int {(x^{2} + 2)}^{- 1 / 2} d x = l n (x + \sqrt{x^{2} + 2}) + C t e$
	Commented by mr W last updated on 28/Jan/20

	$s i r :$ $\frac{d (\ln y^{'})}{d x} = \frac{y^{″}}{y^{'}} \neq \frac{y^{″}}{y}$ $\frac{y^{″}}{y} = - \frac{x}{x^{2} + 2} ⇏ {l n y}^{'} = - \frac{1}{2} l n (x^{2} + 2) + C$
	Commented by abdomathmax last updated on 28/Jan/20

	$n o t c o r r e c t$
	Commented by jagoll last updated on 28/Jan/20

	$how to get it ?$
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