solve y^(′′) −xy =0 by using integr series.


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Question Number 57925 by maxmathsup by imad last updated on 14/Apr/19

$s o l v e y^{^{″}} - x y = 0 b y u s i n g i n t e g r s e r i e s .$
Commented by maxmathsup by imad last updated on 18/Apr/19
$let search develloppable at integr serie let y =Σ_(n=0) ^∞ a_n x^n ⇒ y^′ (x) =Σ_(n=1) ^∞ na_n x^(n−1) and y^(′′) (x) =Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) (e) ⇒Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) −Σ_(n=0) ^∞ a_n x^(n+1) =0 ⇒ Σ_(p=0) ^∞ (p+2)(p+1)a_(p+2) x^p −Σ_(p=1) ^∞ a_(p−1) x^p =0 ⇒2a_2 +Σ_(p=1) ^∞ { (p+1)(p+2)a_(p+2) −a_(p−1) }x^p =0 ⇒a_2 =0 and (p+1)(p+2)a_(p+2) −a_(p−1) =0 ∀ p≥1 ⇒a_(p+2) =(a_(p−1) /((p+1)(p+2))) ∀p≥1 and a_2 =0 ⇒a_(2n+2) =(a_(2n−1) /((2n+1)(2n+2))) and a_(2n+3) =(a_(2n) /((2n+2)(2n+3))) y(x) =Σ_(n=0) ^∞ a_n x^n = Σ_(n=0) ^∞ a_(2n) x^(2n) +Σ_(n=0) ^∞ a_(2n+1) x^(2n+1) =Σ_(n=2) ^∞ (a_(2n−3) /((2n−1)(2n+1))) x^(2n) +a_0 +Σ_(n=1) ^∞ (a_(2n−2) /(2n(2n+1))) x^(2n+1) also we can use that (a_(2n+2) /a_(2n−1) ) =(1/((2n+1)(2n+2))) ...be continued...$
$l e t s e a r c h d e v e l l o p p a b l e a t i n t e g r s e r i e l e t y = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow$ $y^{^{'}} (x) = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1} a n d y^{^{″}} (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ $(e) \Rightarrow \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 0 \Rightarrow$ $\sum_{p = 0}^{\infty} (p + 2) (p + 1) a_{p + 2} x^{p} - \sum_{p = 1}^{\infty} a_{p - 1} x^{p} = 0 \Rightarrow 2 a_{2} + \sum_{p = 1}^{\infty} {(p + 1) (p + 2) a_{p + 2} - a_{p - 1}} x^{p} = 0$ $\Rightarrow a_{2} = 0 a n d (p + 1) (p + 2) a_{p + 2} - a_{p - 1} = 0 \forall p ⩾ 1 \Rightarrow a_{p + 2} = \frac{a_{p - 1}}{(p + 1) (p + 2)}$ $\forall p ⩾ 1 a n d a_{2} = 0 \Rightarrow a_{2 n + 2} = \frac{a_{2 n - 1}}{(2 n + 1) (2 n + 2)} a n d a_{2 n + 3} = \frac{a_{2 n}}{(2 n + 2) (2 n + 3)}$ $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} = \sum_{n = 0}^{\infty} a_{2 n} x^{2 n} + \sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}$ $= \sum_{n = 2}^{\infty} \frac{a_{2 n - 3}}{(2 n - 1) (2 n + 1)} x^{2 n} + a_{0} + \sum_{n = 1}^{\infty} \frac{a_{2 n - 2}}{2 n (2 n + 1)} x^{2 n + 1}$ $a l s o w e c a n u s e t h a t \frac{a_{2 n + 2}}{a_{2 n - 1}} = \frac{1}{(2 n + 1) (2 n + 2)} . . . b e c o n t i n u e d . . .$
	Answered by 121194 last updated on 14/Apr/19
	$y=Σ_(i=0) ^∞ a_i x^i y′=Σ_(i=0) ^∞ a_i ix^(i−1) =Σ_(i=−1) ^∞ a_(i+1) (i+1)x^i =Σ_(i=0) ^∞ a_(i+1) (i+1)x^i y′′=Σ_(i=0) ^∞ a_i i(i−1)x^(i−2) =Σ_(i=−2) ^∞ a_(i+2) (i+2)(i+1)x^i =Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i y′′−xy=0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −xΣ_(i=0) ^∞ a_i x^i =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −Σ_(i=0) ^∞ a_i x^(i+1) =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −Σ_(i=1) ^∞ a_(i−1) x^i =0 2a_2 =0⇒a_2 =0 (i+2)(i+1)a_(i+2) −a_(i−1) =0;i>0 a_(i+2) =(a_(i−1) /((i+2)(i+1)));i>0 a_i =(a_(i−3) /(i(i−1)));i>2 a_0 =c_1 a_1 =c_2 a_2 =0 a_3 =(a_0 /6)=(c_1 /6) a_4 =(a_1 /(12))=(c_2 /(12)) a_5 =(a_2 /(20))=0 a_6 =(a_3 /(30))=(c_1 /(180)) ... y=Σ_(i=0) ^∞ a_i x^i =Σ_(i=0) ^∞ a_(3i) x^(3i) +Σ_(i=0) ^∞ a_(3i+1) x^(3i+1) +Σ_(i=0) ^∞ a_(3i+2) x^(3i+2) =c_1 Σ_(i=0) ^∞ (x^(3i) /(f(i)))+c_2 Σ_(i=0) ^∞ (x^(3i+1) /(g(i))) f(i)= { (1,(i=0)),(((f(i−1))/(3i(3i−1))),(i>0)) :} g(i)= { (1,(i=0)),(((f(i−1))/(3i(3i+1))),(i>0)) :}$
	$y = \underset{i = 0}{\sum^{\infty}} a_{i} x^{i}$ $y^{'} = \underset{i = 0}{\sum^{\infty}} a_{i} {i x}^{i - 1} = \underset{i = - 1}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i} = \underset{i = 0}{\sum^{\infty}} a_{i + 1} (i + 1) x^{i}$ $y^{″} = \underset{i = 0}{\sum^{\infty}} a_{i} i (i - 1) x^{i - 2} = \underset{i = - 2}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} = \underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i}$ $y^{″} - x y = 0$ $\underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} - x \underset{i = 0}{\sum^{\infty}} a_{i} x^{i} = 0$ $\underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} - \underset{i = 0}{\sum^{\infty}} a_{i} x^{i + 1} = 0$ $\underset{i = 0}{\sum^{\infty}} a_{i + 2} (i + 2) (i + 1) x^{i} - \underset{i = 1}{\sum^{\infty}} a_{i - 1} x^{i} = 0$ $2 a_{2} = 0 \Rightarrow a_{2} = 0$ $(i + 2) (i + 1) a_{i + 2} - a_{i - 1} = 0; i > 0$ $a_{i + 2} = \frac{a_{i - 1}}{(i + 2) (i + 1)}; i > 0$ $a_{i} = \frac{a_{i - 3}}{i (i - 1)}; i > 2$ $a_{0} = c_{1}$ $a_{1} = c_{2}$ $a_{2} = 0$ $a_{3} = \frac{a_{0}}{6} = \frac{c_{1}}{6}$ $a_{4} = \frac{a_{1}}{12} = \frac{c_{2}}{12}$ $a_{5} = \frac{a_{2}}{20} = 0$ $a_{6} = \frac{a_{3}}{30} = \frac{c_{1}}{180}$ $. . .$ $y = \underset{i = 0}{\sum^{\infty}} a_{i} x^{i}$ $= \underset{i = 0}{\sum^{\infty}} a_{3 i} x^{3 i} + \underset{i = 0}{\sum^{\infty}} a_{3 i + 1} x^{3 i + 1} + \underset{i = 0}{\sum^{\infty}} a_{3 i + 2} x^{3 i + 2}$ $= c_{1} \underset{i = 0}{\sum^{\infty}} \frac{x^{3 i}}{f (i)} + c_{2} \underset{i = 0}{\sum^{\infty}} \frac{x^{3 i + 1}}{g (i)}$ $f (i) = {\begin{cases} 1 & i = 0 \\ \frac{f (i - 1)}{3 i (3 i - 1)} & i > 0 \end{cases}$ $g (i) = {\begin{cases} 1 & i = 0 \\ \frac{f (i - 1)}{3 i (3 i + 1)} & i > 0 \end{cases}$
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