Question

18 Pocuer series Soution (x2+1) y"+ xy +zx=0, gCO) = 2, y 'CO)=3

Answer 1

Power series 16?+o) y + xy'+ 2xy =0 , 410)=2 = 2 . y'10)=-3 Solution. Given that 122+197' + xy' + 20 Dividing by (x! +1), can be written in standary formas. (2) yu' + x a y + 2x y=0 (22+0) (x2+1) Comparing (2) with yo+Pexy' + &()-o here we have x and pex) = 2 x P(x) = (x2+1) (x2+1) anal PCX) Showing that Q(x) are analytic at x=0 so x=0 an ordinary point. Therefore to solve cus take power series. y = acotcxt C₂ 2 sc? +C3? + I cnxn (3) nao Differentiating 37 twice in succession wigit: 'x' we get n-1 y'= Incoza- swi and ya 2 n(n-1) Cnxn-2 n=2 4) Substituing the value of y, y' and you in (l) we get

Cnrix" (be?+1) I min-1)( xn-2 + x nenxenal + 2x enah n22 nol nio m-2 nen-1) CnX cna + + W nenan a nen-1) cnx n=2 n=2 n=1 + 2 Z en anti co n=d a 5 nen-1) ensen in+2779 +1) Cntza m2 nal =0 n=1 02 18 222 + (669 +4, +2 Co) +) { min-13cm + (n +2) +1) n=2 Cmt 2 + m Cn + 2 C - { x" =. (5) Sinie (5) is an identity, equating the constant term. powers of a to zero we get and the coefficients of various and 6C3+ C + 200 =0 22=0 [C2=0 from (3) y = Co taiatezze? +

ylo)Co = 2 therefor! Co-2 intial condiction given y (0)=2 and y'= C, +264 x + 363x2+ y 10) = 4, 5.3 C =3 663 +9, +2 Co= .663 t-3 + 2x2 = 0 663 +1 -0 C3-16 minal) Conten+2)(n+1) Cast2 them + 2 Cn-l E and --> 02 Con + (+2)into) Cmt2 + 2 Chai = 0 Cnta 2 Con- =on² cn in+2)(n+1) 12? cnt 2 Cn-1) Conta for all m), 2. (6) (n+2)(n+1) The above relation (6) is Korown as gelation recurrence Putting n=2 in (6) CA = - (462 +264) 12 C4 =-2co 12 - 28-3 o C4= = 12 2

C4 ין 2 1 Putting n=3 cs =-(9Cg+26) 20 (9x-1 + 2x0) - B2 20 = 3 2. 20 Cg 17 3 3 40 P of Co, C, ,C2, C3, C4, Cg etc in putting the ralue egn (39 we get y= 2+(-3 Due to.be + 33+ 34+3.25 + 40 3 y = 2-30 2 -39 - 4x3+ + 24 + 3 35 6 2 40 required Solution near x=0 which is the