Question

Find two power series solutions of the given differential equation about the ordinary point x = 0. y′′ − 4xy′ + y = 0

Find two power series solutions of the given differential equation about the ordinary point x = 0. y!' - 4xy' + y = 0 Step 1 We are asked to find two power series solutions to the following homogenous linear second-order differential equation. y" - 4xy' + y = 0 By Theorem 6.2.1, we know two such solutions exist about the ordinary point x = 0. In fact, the differential equation has no singular points as each of the coefficient functions are very simple) polynomials. Therefore, they are trivially represented as power series at any point and are therefore analytic. This means the power series solution will converge for 1x1 < 0. Let y = n = 0 solve for two sets of cn to find the two solutions. to cox". We substitute this series and its derivatives into the given differential equation. We will The differential equation includes non-zero terms for the first and second derivative of y. These derivatives always have the same form, so we refer to equation (1) of section 6.1. y' = ☺ canxa - 1 n = 1 y" = n(n - 1)x7 - 2 n = 2 Substitute the power series into the given differential equation. y" - 4xy' + y = Š (c, n(n − 1)x9 2) - 4 (C (CMX") n = 2 n=1 n = 0 Submit Skip (you cannot come back)

Answer 1

(1) Ans: Given yu-axy'ty=0 solution of 1). let, yeah ท be << -| d. Enent Now, -| X 1-2 H - 2 (ก-1) 72 x Therefore, Zn(n-l), n-2. n- - 4 need - 20 t 2 en 2 -2. 220 ท-2 >>.(n-1) -42 ne6 +2 C% +2 ,3 20. -L 것계 h0 x -2. 1-2 -4 (n-je en 1-2 1. -0 2. 552 1-2 -> (n-1) + 2 2 1P3 * (n-1]e 3 -45 (ท-202-2 2,282 test en h3 ท-2 1. 22 L h=3 1-2 1 L-2 2 0 a * (2e++.)+2 (n-1)* 4(n-2) 1-2 “ก-21 0 >(26+e.)xn-1)* 4-191-27-30 -> (2e, t C% ) +2 19-10° +(9-47) 1-213 1-2 0 0. 23 s 0 2.e, tea 20. => 2 - - 24 ONd (7-1Jet(94) 1-2-0. (4n-9บ (-) (1) en Cn-2

putting a putting 2=3 in (ii) then we have ez 12-9) e 3(3-1) = £9 2=4 in ii) then we have (16-9) 4(4-1) - 73 (-{co) eA وع 3-7 zas putting 24 22=5 . in is then we have eg 515-11 =(20-9203 12 20 (49) 2609 putting nag in my then we have, CG (24-9) CA 6 (6-1) 15. .(- Z 30 ) 24 2 Co. (il, then we have, putting 48 n=7 in lil (28-9) 7 (7-1) eq 19: Les 1409) . 42 - 209 168 9 putting n=8 in it, then we have, eg (32-9) 8.08-1) 6 23 7 - 56 161 - 2688 Co

and so on. n n0 yel) - Legal 2 eg te 2 te x 7 ez a 4 c4 24+ esateabregat tegyet =gte, 2-16x +49x² / 26x1+hear 209 7. 161 Beace t 48 7 16a7 1680 2688 8x8 = (1-11 2 2 2 2 2 2A-7 x 24 48 2688 X6 167 x ) --- where +9(+423 +46 XS+ 20%. ++ %.es are arbitrary constants Required, power series sotution of ing is .: given b iii)