Question

Given the DE: y"-(x+1)y'-y=0 use it to answer the following:

a) Find the singular point(s), if any, and if lower bound for the radius of convergence for a power series solution about the ordinary points x=0

b)The recurrence relation Hint: It will be a 3-term recurrence relation

c)Give the first four non-zero terms of each of the two linearly independent power series solutions near the ordinary point x=0

Answer 1

Solution : Siven, y" - JOC+ 1)y! -y=0 (1 Now compare with y"+ P(x):Y'+916) Y=0. → D (OC) = - (C+1) & 26%) = -1 since Play & Q (2) are analytic yan every LER. so , ey" y_ (alt) y'-y=0 has no singular point. & Now let y(x) = { an sc aboutt-o, te a series salm of equ li then radius of convergence your this power series salm be pateo. - ie rower series solm about do=o converges #o</21 <co m- 6) Y'(x) = į naman- & ylle) = ě n(n-1) anon2 n=1 n= put these into li), - n-2 s nen-1) and I M-1 tile nana n=2 00 { angin=o n=0 . - =) 00 in-2 į non-1) ans" n=2 me n į nana n=1 in-1 į nanxe" - nat ě anon=0 n=o Now replace n-a by n into Ist series suma - n-1 by n into Iord series sum, @ In+2)(n+1) Ant2 2" name" 2 (malam pertama anx?=o = net noot n=o no

= P2280 - Q4 seº-2020+ [lo+2)(n+1 Jan 72-(n+1) An- lon+2 )An+1]&#07 que compare the coefficients of power's of e both sides, PA220 Q 270-008°=0 =) Az = gf [94+60] f ((n +2 ) Ant2 - Am - Antza m+)]>"20 * nzz 7 (n+2) Anta-an-antt =0 &nst 80 is given by The recurrence relation anta= an tanti (n+29 anso Now, n © e to anta 22 = 1104+20) = a1 +02= Pattat +90 – 301+00 3 3x3 a 4= A 2 +93 - 192+90 H3 at +90)/2x3 4 = 394 + 390 +394+ho sa Px3x4 a RX3 +4 ao 2x3x4 3 as = a 3 +94 - 13Q1+90) +6Q+ +420) 2x3x4 5 2x3 = 12att 400+692 +420 – 1827+ 890 2x 3X 4X 5 51 4 û 6 = a 4+9s (6U1+4 slo) + 11897 +840). 41 s! =

= BOUL+20610 +189 +800 4894 + 2800 6! 61 § 27 = ag+Q6 - 18Q+ +880 + 4894 +2890 - S! - 61 7. = 10821+ 4890 + 4897 +2800 = 15691+7690 7! 7! So, y(t) = Qo tape + Az x + 234*+--- = a +99 $ + (044a,3 + 30 +0.) 4 6 2 1+4 - 4 a at 1+ 4 ' 41 2! 3! +(1894+800) 25+ (4 gat + 2800 )x+ (15697+769017 - 5 - 61 L -- = 00[1+ ***** ****,2448454---] * 1 Lữ +3, +3+4 +4 --] of then ao & ay be two hago non-zero , arbitary constant them power series saln are given to 4160 = 4 + g 2 + + *+-- * 1971x) = ge+ i + x + x + ---