Question

4. Given that x =0 is a regular singular point of the given differential equation, show that the indicial roots of the singularity do not differ by an integer. Use the method of Frobenius to obtain to linearly independent series solutions about x = 0. Form the general solution on (0, 0) kxy” – (2x + 3)y' + y = 0

Answer 1

2x4" - (2x + 3) + 20 seguloa singular point assume the solution y = { chach n:o whene Coto differentiate term by term n+-1 y': { [n+r) Cn (x-2) no nir-2 ya E (nes) (nta-1) cn (x-xo) L nuo Substitute © © in o Don't forget that Xo = 0 n+2-1 nis-2 2x{ (hts) (n+si-1) cn (x) - 2x{ (hts) cn (x) no no E nir Cnx + teo (nts) cn achts- nco про n+9-1 - 2E (nter) in achts 2€ (nts) (nts-i) Cn x' no n:o n+r-1 { Cn acht -3E (n+r) Cn xh nao hoo

we shall re-write the first and third Summation so that each summation will have the exponent a ntr X ih Consider first Summation ntr-I 28 (hta) (ntr-) en x nao put n-i=m h=mti mer 2 m++) (mar) Cmtix mor! 8 W 2 (a + s +1) (nts) Cnti achth + no-1 Consida 3 3rd summation - § (nts) en achts- no pot n-1=m oo nant -3 E (m+8+) Cmti xh mih m=-1 -35 (n+a+1) Cnti a ntr aans-1

Page No: -20 2E hta Th+r+1)(n+a) Chrixch a Cnts) Chanta no- ኣ = 4 00 -3E In +974) Chrisch na ner .cna +1 huo 9-1 2016-1) Cox +2€ (n+r+1)(n+8) Cntia" no -24 (nts) Cn x n+r nco -3 r Coari -38 (n+8+1) Cnti ght? + no + [ Cn x hth no Equating Oljep to 91 2 2010 the coefficient of the lowest power of (oto ہ= و3 - (-وو2 ELHALANAS) 2 r².25-35 = 0 29?-5 si = 0 r(ar-s=o r=0 er = 2

hir Comparing coefficient of coefficient of power of ac 2 (n+r+1) (nta) Cnti o 2 (n+r) cn -3 (n+9+) Chal t CA O Mto Conti Cntr41) (2(hts) -3) + Cn (1-2 (n+8)= -cn (1-2(n+8)) = = Cnti (ntat (26n+3) - 3

(87 nzo Chti - Cn (1-2(n+8)) (n+3+1) (26m+9)-3) pot no G to pot r=o Cuti Cn (1-2n) (n+) (2n-3) put ทะ 0) C, -lo Co 3 -3 put n=1 C2= ca (+1) -C7 I co 2 3 Co 2 (+1) 2 6

put n: 2 (3= -C2 (-3) (2=-co 13) (1). 3 6 Put n=3 -631-5)$13=57-6 Cu = -3 (1-6) 14) (3) 12 1216 co 32 we get the solution Y = Cotx'+ C2 x² +63x² Cox² Cot Cool st -Co 2 26 3 6. Now Jet are sa Chri & hao no Cn (1-2 (nt)) (n+ul) (2/n+$}-3) Cn (2nty th+) ( 2n+2) Cn (n+2) (n+3)( n+y) n=o 26 2 C = Co (2) 12 / 2x' 260 = 46 7

= ) С, c, (3) 3C, 3. Су Ч.. 2. 3 4 (gy2) 1.2 (+2) (+1) Ч.С. 2) - 2 (с(4) (22) (2+) ЧС, (4). 3 ЧС, 2. 33 ВС, 33. 3 ч С. 33 2) 32 с. 69 3 Now solution ГА 는 С, x + c x +cix G x² + Colt Cox X-T чь хя 32. Се т. x 2 + 4 Со 2 2) 193 general soll - Ч С Ч + C2 Ч..