Question

As. Find a power-series expansion the General solution to the differentar equation yll-Sxy' tyo

Answer 1

Let the power seves expansion g(n) = [annn n=0 be the general solution of the given differential equation. Now, differenti itiating wort n, we geti n=1 y'(x) = {nanan again diff n(n-1) an xn-2 wort a we have y" (n) n=2

Now the transformed differential eqn n-2 Ź n(n-1) anni - 5 n E hannn na2 + annn =0 n=0 E n(n-1) an yn-2 8W - 01 nannn ↑ na 2 n=1 + { anno no As we know [ n(n-1) anni { (+2)(n+1) antz n-2 n=0 Now the differential equation → E (n+) (nn) ant2 nn - 5 hanno + Ž n=0 n=1 annh h=0

+60+2) (0+1) aq n° + [n+2)(n+1) 90+2 yn n=1 -52 nannn nanna taon+ ܐܶ܀ anna=o n=1 n=1 → E [{n+2) (n+1) ant2 - 5 nant an Inn n=1 + (292 +90) = 0 Now comparing Rides, we the coeffivent of both get (24, + 4) = 0 → a= and (n+2)(n+1) ant? - 5 nantano, = ao 90 2 ns1

i na = (n+2)(n+1) antz - (55-1) an = 0 9n+2 = 5n-1) ann21 (n+2)(n+1) Put n=1 5.11 1 44 (1+2)(1+1) 4 3.2 34 n=2, an 16-1 = 92 (2+2)(2+1) 11 9 az = 9 4. 3 90 2 4.3 g 11 90 2.3.4

If n is even. then (0-2) is also even then 5h-1 anta (1 +2)(n+1) 5 (1–2)-1.90-2 (n+2)(n+i) (n=2+2)(1+2+1) 5h-1 5 since on = 5(n-2)-1 an-2 ((A-2) +2) (1+2+1)

(n+2)(n+i) ((1-2)+2)(0-24) (2+2)(2+1) ant2 = le 51-1 5(n-2)-1 5.2-1 . ag If h is odd. then 1-2 is odot: then an+2= 5n-1 5(n-2) - 1 (6+2)(n+1) ((1-2) +2)(n-2+1) 5.1-1 (1+2)(1+1) 91 Thus Y(n)= Ï annh h30 8y = a zno, tanto aan nen hoo heb

Now, aanti nalt annel 92n+1+2-2 (5 (2n-1) -1) (6 (2003)-1) ( 20–4 +2) (2n-1 +1) (20-3+2) (20-3+1) le 92n-1) +2= 51-1 (1+2)(1+1)

Now aan (21-2+ (ต-1)+) 1 .99 5 โ2(-]- ((n-1) +2) (2(ต-) +1) 52 - 1 (1+1) (241) Thus th *N aan + 12h+) () = 9thy h= 0 TT 5 (12-1)-1) (5 (2n-3)-1) (ใก- 42)(In-1 41) (ขต 42)(16-341) h= 0 5-1- หh+ (142)((+1) 59-| + h-s A 14 ((n-1)-1) (2 (F-1) 11)(1 (-1)+) (-) (14) 5 ((n-1) -| ((n-1) +2)((n-1)++) (5 ((n-1) -1) ((91) +) ((ตะ+11 ” เจ” 2. 3-1 5 4 \(1) - hะง 9 4 4.3

for a given a_{​​​​​​0}  and a_{​​​​​​1} .