let f(x)=e^(−x^2 ) ∫_0 ^x e^t^2 dt 1) find a d.e verified by f 2) developpf at integr serie.


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Question Number 34699 by abdo imad last updated on 10/May/18

$l e t f (x) = e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t$ $1) f i n d a d . e v e r i f i e d b y f$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by math khazana by abdo last updated on 11/May/18
$we have f^′ (x)= −2x e^(−x^2 ) ∫_0 ^x e^t^2 dt +e^(−x^2 ) e^x^2 =−2xf(x) +1 so f issolution of the d.e. y^′ =−2xy +1 2) let put f(x)=Σ_(n=0) ^∞ a_n x^n ⇒f^′ (x)= Σ_(n=1) ^∞ na_n x^(n−1) f^′ +2xf −1=0 ⇔ Σ_(n=1) ^∞ n a_n x^(n−1) +2x Σ_(n=0) ^∞ a_n x^n −1 =0 ⇔ Σ_(n=0) ^∞ (n+1)a_(n+1) x^n + 2Σ_(n=0) ^∞ a_n x^(n+1) −1 =0⇔ Σ_(n=0) ^∞ (n+1)a_(n+1) x^n +2 Σ_(n=1) ^∞ a_(n−1) x^n −1 =0⇔ a_1 + Σ_(n=1) ^∞ {(n+1)a_(n+1) +2 a_(n−1) }x^n −1=0 ⇔ a_(1 ) =1 and ∀n≥1 (n+1)a_(n+1) +2 a_(n−1) =0 ⇒ (n+1)a_(n+1) = −2 a_(n−1) ⇒ a_(n+1) =((−2)/(n+1)) a_(n−1) ⇒ a_(2n+1) = ((−2)/(2n+1)) a_(2n−1) and a_(2n) =−(2/(2n)) a_(2n−2) =((−1)/n) a_(2n−2) Π_(k=1) ^n a_(2k+1) = (−2)^(n+1) ((Π_(k=1) ^n a_(2k−1) )/(Π_(k=0) ^n (2k+1))) .a_3 .a_5 .....a_(2n+1) = (((−2)^(n+1) )/(1.3.5....(2n+1))) a_1 .a_3 ...a_(2n−1) ⇒ a_(2n+1) = (((−2)^(n+1) )/(1.3.5.....(2n+1))) also we have Π_(k=1) ^n a_(2k) = (((−1)^n )/(n!)) Π_(k=1) ^n a_(2k−2) ⇒ a_2 .a_4 .....a_(2n) = (((−1)^n )/(n!)) a_0 a_2 ....a_(2n−2) ⇒ a_(2n) = (((−1)^n )/(n!)) a_0 f(x) = Σ_(n=0) ^∞ a_(2n) x^(2n) +Σ_(n=0) ^∞ a_(2n+1) x^(2n+1) = Σ_(n=0) ^∞ (((−1)^n )/(n!))a_0 x^(2n) +Σ_(n=0) ^∞ (((−2)^(n+1) )/(1.3.5...(2n+1))) x^(2n+1)$
$w e h a v e f^{^{'}} (x) = - 2 x e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t + e^{- x^{2}} e^{x^{2}}$ $= - 2 x f (x) + 1 s o f i s s o l u t i o n o f t h e d . e .$ $y^{^{'}} = - 2 x y + 1$ $2) l e t p u t f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow f^{^{'}} (x) = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1}$ $f^{^{'}} + 2 x f - 1 = 0 \Leftrightarrow \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ $+ 2 x \sum_{n = 0}^{\infty} a_{n} x^{n} - 1 = 0 \Leftrightarrow$ $\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + 2 \sum_{n = 0}^{\infty} a_{n} x^{n + 1} - 1 = 0 \Leftrightarrow$ $\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + 2 \sum_{n = 1}^{\infty} a_{n - 1} x^{n} - 1 = 0 \Leftrightarrow$ $a_{1} + \sum_{n = 1}^{\infty} {(n + 1) a_{n + 1} + 2 a_{n - 1}} x^{n} - 1 = 0 \Leftrightarrow$ $a_{1} = 1 a n d \forall n ⩾ 1 (n + 1) a_{n + 1} + 2 a_{n - 1} = 0 \Rightarrow$ $(n + 1) a_{n + 1} = - 2 a_{n - 1} \Rightarrow a_{n + 1} = \frac{- 2}{n + 1} a_{n - 1} \Rightarrow$ $a_{2 n + 1} = \frac{- 2}{2 n + 1} a_{2 n - 1} a n d a_{2 n} = - \frac{2}{2 n} a_{2 n - 2} = \frac{- 1}{n} a_{2 n - 2}$ $\prod_{k = 1}^{n} a_{2 k + 1} = {(- 2)}^{n + 1} \frac{\prod_{k = 1}^{n} a_{2 k - 1}}{\prod_{k = 0}^{n} (2 k + 1)}$ $. a_{3} . a_{5} . . . . . a_{2 n + 1} = \frac{{(- 2)}^{n + 1}}{1.3.5 . . . . (2 n + 1)} a_{1} . a_{3} . . . a_{2 n - 1}$ $\Rightarrow a_{2 n + 1} = \frac{{(- 2)}^{n + 1}}{1.3.5 . . . . . (2 n + 1)} a l s o w e h a v e$ $\prod_{k = 1}^{n} a_{2 k} = \frac{{(- 1)}^{n}}{n!} \prod_{k = 1}^{n} a_{2 k - 2} \Rightarrow$ $a_{2} . a_{4} . . . . . a_{2 n} = \frac{{(- 1)}^{n}}{n!} a_{0} a_{2} . . . . a_{2 n - 2} \Rightarrow$ $a_{2 n} = \frac{{(- 1)}^{n}}{n!} a_{0}$ $f (x) = \sum_{n = 0}^{\infty} a_{2 n} x^{2 n} + \sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} a_{0} x^{2 n} + \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n + 1}}{1.3.5 . . . (2 n + 1)} x^{2 n + 1}$
Commented by math khazana by abdo last updated on 11/May/18

$w e h a v e f (x) = e^{- x^{2}} \int_{0}^{x} e^{t^{2}} d t \Rightarrow$ $f (- x) = e^{- x^{2}} \int_{0}^{- x} e^{t^{2}} d t$ $=_{t = - u} e^{- x^{2}} \int_{o}^{x} e^{u^{2}} (- d u) = - e^{x^{2}} \int_{0}^{x} e^{u^{2}} d u = - f (x)$ $s o f i s o d d \Rightarrow a_{2 n} = 0 \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} \frac{{(- 2)}^{n + 1}}{1.3.5 . . . . (2 n + 1)} x^{2 n + 1} .$
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