Question

Use the Frobenius method to find the general solution near x = 0 of the hypergeometric equation x(1 - x)y" + [C - (A + B +1)x]y' - ABy = 0 where A and B are any real numbers, and C is any real nonintegral number.

Answer 1

The given Hypergeometric equ is *(1-x) y" + [C- (A+B+1)x]Y' - ABY=0 where A and B are real numbers and is any real non integral no. Let Po (n) = -AB Pi(n) = C-(A+B+1) P2(X) = x(1-~) Then 2 > 0 1- o P2 (0) = Pall)=0 Hence, a=0 and n=1 are singular points Let's start with n=o. Lin x-a) P, (x) in 6e-o) [c- (A+B+Dx] Pan) x (1-21) Lin & [c- (A+B+1)x] x (1) Lin (2-a)? Pole) Lin (X-0² (-AB) - Linn?(-AB) P2 (2) x(1-2) x(1-2) 210 N-a 70-> 220 Hence Both clinit enist and a zo is a regular Singular point. Therefore, we assume the seel" of ilhe form y = Har xeta r=0 with ao to. Hence, diff. w. s. tix y'a Ž an (2+) 3216-19 co

༢ +-2 s" –(༨༨) (༡༨༤༥) > =9 using + ༢ -༠ +4 - 7 - as》བ ༩༩ ཏུ=o = =5 ༼ ༥ འི -༽ » - Y 2 ༢ =5 ཉི=c ༢ནི། + ༢༠ ༢༤༩ ༢ ༤ ༢༥༢ & in given ean ཉ7(- ༧), ( ༢ +8 (༡ +༩-) ག - [c-(ཀཆ ༤ ༥)x] a, (+4) ) ༧》༠ (ཀ ཏ༥) (༢༠༩ པད) ༼༣་༥ - , ་Za, (༢ )(2༨༢-༨） cZa, (༢ + ༢)? ཧ -། (a + 13 + ད)7ོa-(༣༨༣) - A ད、 2。 ར ངོa (༣ ༨ ༨)(༡༥ པད) ༣༨ - _ ༼བ, (༩ ༨༤) (༩༥༩ བད)ཌ་ ངོ (༢༨ ༢) ༡༩ ར ལ - – (a རྡ 3ཤད), (༣༨ ༣) ༡ ཉན -23 ོ ༡༣ +༩ Here will need all powers to be the same, equal to e+c-1, the smallest power Hence, °༣ (༡༥༢) (» +d-ད) -2° arol （A +） (54 ནི བ༢) +༩༼བ, (t +༩) ༼༡ 4 -- (0༥ 8ཤད)ོ- (+གིས) ༼ ༤ ། ༢ o ༢ ༤༠ ༢ =o ཀཾ=5 དུསཞཔ| $+4- ཤ= ། ཧི =4 ཏུ=o ༤༠༠ a. R+d -། ༢ +d -| Thus, isolating the first term of the sums starting from all get. ༢༠(༩-1) zh- +Cqའི ༤- ，, a, (%+ a) （༢+4 -1) -Z_4-༥ (༣ ས 1 ）( +4-1) ཎ ཏ༥-། +cZo, (3+༢) ཀཾ ན པ| ༢ -། 원의 ཏེ།

Eta-1 - (A+B+1) 94(=1) 38*44_ AB Ž qę m? r=1 Now from the linear independent of all powers of x, that is, of the function I, n, n? etc. the coifficients of ze vanish you all k. Hence, from the first term, we have aola (0-1) + Ca)=o which is the indicial eq". Since as to we have a (2-1)+ca=o a (2-1+6)=0 di = 0 Also from the rest of the terms, we have [(x ta)(4+2-1) + C($+d)]9x + [-(8+2+1) (2 +a-2) – (1+A+B) (&+=1) - AB]2=C- x2 = 1-C Hence ar = arol (+2-1) (r+2-2) + (1+A+B) (2+0-1) +AB -921 d+Q) (r +0-1) + C(r+2) (2+2-1) (Eta+ A+B-1) + AB (d+Q) (r + 2 +C4) But (92 +0-1)(8 + +A+B-1) + A3 = (2+0-0(+0+ A-1) + 2+2-1) BT AB = 4+2-1) (& +2+ A-1) at B[s+a+A-) Hence we get the recurrence relation (de + d + A-1) (2+2 +-1) ar = (& td) (& +2+(-1) -aqui for {>

qo Now, simplify this relation by giving as in terms of as instead of art a, C) (2+A) (2+B) (2+1) (2+0 92 - A+ A+) (2+B+1 a, (Q+A+1)(Q+A) (+B+1) (+B) (@+2) (a +C+1) & +2)(2+1) (d+C+1)(2+0) (d+A), (&+B)2 (2+1), (Ql+C), 9. ܘܘ 93 - d+A+2)(Q+B+2) DA), (Q+A+2)(+B), (+3+2) &+3) (+C+2) (x+3)+1) (+) (d+C+2) (Q+A), Q+B) (+1)(+c)? do 9= (2+A)ų (2+B)ę for den (x+1), (X+C), Hence the solution y=aoE +A), (+B)! r=0 (2+1), (6+C)ę arta