Question

A power series solution about x=0 of the differential equation y"-y=0

answers

3 H, - A EWE BW H22 1 - 2 22 ปอ: 388 1 SWPWE PW PL SWE 2, 2 2พะ 1) 9 - 20 4

Answer 1

Solving the power series solution we get
The given differential equation is ylyzo comparing with bolny"+ Pilny't P (n)y=0 we get Polny = 1 => Poln) to at nao >n=0 is an ordinary point of the differential equation. Let us consider the series solution o ay y = anun no ie, y=dotan + a₂2²+azut + annat + nanana :y' = a +291 +39 u²t y" = 29 +3.2a3ht »y" = 29₂ +69xt putting the value of y + n(n-1) anun-2 + n(n-1) ankh-2 yu ini we get 12.0, +6031+ ... (n-1) nanah]-[antan tqu²+azua+ -t annat -]=0 Equating the co-efficent of no (lowest degree term) we get 20-do-o 292= ao من 3 az = 2

Equating the co-efficent of n we get, 693 -c=0 693 =ą az G Equating the co-efficent of an we get (n+1)(n+2) antz - an so an ant2 = (n+1) (n+2) put n=2 in lij we get az a2+2 (2+1) (2+2) -> au (314) - au a2 12 ao aza au 1 12 Iz A > aq ao 24 put n=3 in (iij we get Az+2 аз (3+1)(3+2)

az as (6 (5) az als an * a5 1 20 [ azzal y as 120 put n=4 in (iii) we get an (4+1) (472) a4+2 Ox4 ( 516) au > a= 30 ao co au = 1. 30 [ 24 24 ao » a6 720 putting the value of ao, a. au incij we get y = ao tan & co u² + an w + ao net 24 120 a + ao 720 y = ao .] -- 1 +24² + 1/4 n4 + 1 1/2 n't tag [n+ 1 + + .] 120

22n > y = 20 5 2nti su ta (29 2n+1)! n=0 n=c .: The 3rd option is correct.