Q1)find tow power series solutions of the given D.E about x=0 y^(′′) −2xy^′ +y=0 Q2)use the power series method to solve the given intial value problem y^(′′) −2xy^′ +8y=0 y(0)=3y^′ (0)=0


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Question Number 89624 by M±th+et£s last updated on 18/Apr/20

$Q 1) f i n d t o w p o w e r s e r i e s s o l u t i o n s o f t h e$ $g i v e n D . E a b o u t x = 0$ $y^{^{″}} - 2 {x y}^{^{'}} + y = 0$ $Q 2) u s e t h e p o w e r s e r i e s m e t h o d t o s o l v e t h e$ $g i v e n i n t i a l v a l u e p r o b l e m$ $y^{^{″}} - 2 {x y}^{^{'}} + 8 y = 0$ $y (0) = 3 y^{^{'}} (0) = 0$
Commented by mathmax by abdo last updated on 18/Apr/20
$y^(′′) −2xy^′ +y =0 let y =Σ_(n=0) ^∞ a_n x^n ⇒y^′ =Σ_(n=1) ^∞ na_n x^(n−1) =Σ_(n=0) ^∞ (n+1)a_(n+1) x^n and y^((2)) =Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) =Σ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^n (e)⇒Σ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^n −2Σ_(n=0) ^∞ (n+1)a_(n+1) x^(n+1) +Σ_(n=0) ^∞ a_n x^n =0 ⇒ Σ_(n=0) ^∞ (n+1)(n+2)a_(n+2) x^n −2Σ_(n=1) ^∞ na_n x^n +Σ_(n=0) ^∞ a_n x^n =0 ⇒2a_2 +Σ_(n=1) ^∞ {(n+1)(n+2)a_(n+2) +(1−2n)a_n }x^n =0 ⇒ { ((a_2 =0)),(((n+1)(n+2)a_(n+2) =(2n−1)a_n ∀n≥1 ⇒)) :} { ((a_2 =0)),((a_(n+2) =((2n−1)/((n+1)(n+2)))a_n ∀n≥1)) :} a_(2n+2) =((4n−1)/((2n+1)(2n+2)))a_(2n) ⇒Π_(k=2) ^n (a_(2k+2) /a_(2k) ) =Π_(k=2) ^n ((4k−1)/(2(2k+1)(k+1))) (a_6 /a_4 )......(a_(2n) /a_(2n−2) ).(a_(2n+2) /a_(2n) ) =(1/2^n )Π_(k=1) ^n ((4k−1)/((k+1)(2k+1))) ⇒ a_(2n+2) =(a_4 /2^n )((3/(2.3))×(7/(3.5))×((11)/(4×7))×.....((4n−1)/((n+1)(2n+1)))) we follow the same way to find a_(2n+1) ...$
$y^{^{″}} - 2 {x y}^{^{'}} + y = 0 l e t y = \sum_{n = 0}^{\infty} a_{n} x^{n} \Rightarrow y^{^{'}} = \sum_{n = 1}^{\infty} {n a}_{n} x^{n - 1} = \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n}$ $a n d y^{(2)} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n}$ $(e) \Rightarrow \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - 2 \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n + 1}$ $+ \sum_{n = 0}^{\infty} a_{n} x^{n} = 0 \Rightarrow$ $\sum_{n = 0}^{\infty} (n + 1) (n + 2) a_{n + 2} x^{n} - 2 \sum_{n = 1}^{\infty} {n a}_{n} x^{n} + \sum_{n = 0}^{\infty} a_{n} x^{n} = 0$ $\Rightarrow 2 a_{2} + \sum_{n = 1}^{\infty} {(n + 1) (n + 2) a_{n + 2} + (1 - 2 n) a_{n}} x^{n} = 0 \Rightarrow$ ${\begin{cases} a_{2} = 0 \\ (n + 1) (n + 2) a_{n + 2} = (2 n - 1) a_{n} \forall n ⩾ 1 \Rightarrow \end{cases}$ ${\begin{cases} a_{2} = 0 \\ a_{n + 2} = \frac{2 n - 1}{(n + 1) (n + 2)} a_{n} \forall n ⩾ 1 \end{cases}$ $a_{2 n + 2} = \frac{4 n - 1}{(2 n + 1) (2 n + 2)} a_{2 n} \Rightarrow \prod_{k = 2}^{n} \frac{a_{2 k + 2}}{a_{2 k}} = \prod_{k = 2}^{n} \frac{4 k - 1}{2 (2 k + 1) (k + 1)}$ $\frac{a_{6}}{a_{4}} . . . . . . \frac{a_{2 n}}{a_{2 n - 2}} . \frac{a_{2 n + 2}}{a_{2 n}} = \frac{1}{2^{n}} \prod_{k = 1}^{n} \frac{4 k - 1}{(k + 1) (2 k + 1)} \Rightarrow$ $a_{2 n + 2} = \frac{a_{4}}{2^{n}} (\frac{3}{2.3} \times \frac{7}{3.5} \times \frac{11}{4 \times 7} \times . . . . . \frac{4 n - 1}{(n + 1) (2 n + 1)})$ $w e f o l l o w t h e s a m e w a y t o f i n d a_{2 n + 1} . . .$
Commented by M±th+et£s last updated on 18/Apr/20

$t h a n x s i r a n d w h a t a b o u t Q 2$
Commented by mathmax by abdo last updated on 18/Apr/20

$a d d t h e i n i t a l c o n d i t i o n . . .$
	Answered by ajfour last updated on 18/Apr/20

	$y = c_{0} + c_{1} x + c_{2} x^{2} + c_{3} x^{3} + . . . c_{n} x^{n}$ $y^{'} = c_{1} + 2 c_{2} x + 3 c_{3} x^{2} + . . . . + {n c}_{n} x^{n - 1}$ $y^{″} = 2 c_{2} + 2.3 c_{3} x + . . . . + n (n - 1) c_{n} x^{n - 2}$ $n (n - 1) c_{n} - 2 (n - 2) c_{n - 2} + c_{n - 2} = 0$ $(n + 1) (n + 2) c_{n + 2} = (2 n - 1) c_{n}$ $c_{n + 2} = \frac{2 n - 1}{(n + 1) (n + 2)} c_{n}$ $2 c_{2} = c_{0}, c_{3} = \frac{c_{1}}{6}$
	Commented by M±th+et£s last updated on 18/Apr/20

	$t h a n k y o u s i r$
	Commented by M±th+et£s last updated on 18/Apr/20

	$s i r c a n y o u h e l p w i t h Q_{2} a n d t h a n k y o u$
	Commented by ajfour last updated on 18/Apr/20

	$d o n t m u c h r e m e m b e r a l l t h e s e,$ $i t s a l l t h e r e i n E r w i n K r e y z i g .$
	Commented by M±th+et£s last updated on 18/Apr/20

	$o k s i r t h a n k y o u$
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