Question

10. (4 pts) In this exercise we consider finding the first five coefficients in the series solution of the first order linear initial value problem (x2 +1)y" – 6y = 0 subject to the initial condition y(0) = 3, y'(0) = 3. Since the equation has an ordinary pts at x = 0 and it has a power series solution in the form y = {cnt" no (1) Insert the formal power series into the differential equation and derive the recurrence relation Cn-2 for n=2,3,... The solution to this initial value problem can be written in the form y(x) = coy(x) +C1y2(x) where co and C are determined from the initial conditions. The function yı(x) is an even function and y2(x) is an odd function. For this example, from the initial conditions, we have co =_ and en =- n=1 The function yı (x) is an infinite series yı (x) = 1+ anx2" NOTE note that the constant co has been fac- tored out. (2) Use the recurrence relation to find the first few coefficients of the infinite series ai=_. 22 ,43=_, 04=_NOTE note that the constant co has been factored out. Finally the polynomial y2(x) = NOTE The function y(x) is an odd degree polynomial with first term x. In other words, note that the constant c has been factored out.

Answer 1

0 nel erere cele in + 1) y" - by=0 y" - y = comparing with y" of Pendy't only o P(n) = 0 1 901) = nel the's showing that ears and g(x) are analytic at is an ordinary point. y 0 have so I cnan no 00 n-2 y'- Imani a xnol y" - nin-1cn. n = 2 value in we get putting this (x + 1) E nini) chan-2 6E (nan n = 2 n-o O I min-1) cnan + { non-1) anan-2 2-6 Einuno =0 ma2 n - 2 00 20 n=0 M2 no = I nen-1) (non & I (n+2) (2.1) Cnezan 6 Ž corn=o tput nanta in the second series ] o equating the coefficient of ☺ we get for n=0, (012) (0+1) C2 - 66020 202 – 60o = 0 - C2 = 3cc m 21 ) (1+2)(1+1)C, 12-6ci - 0 EC3 - 64 -0. - c3

min-1) + (n + 2)(nellen 2-6 (n=0 nr.2 chea ch (n + 2) (n + 1) (n +2= benen(n-1) cn (on nen- 122 nez) inti) 6-(0-2)(n-2 - 1.) (n-2 + 2)(n- 221) n72 C- (-2)(n-3) en-2 I onor n(n-1) on=2,3, n. ст (n-2 1 = 2 in @ ca Vi 6- 2 4.3 -C2= 2 4 3 = 1/2 C2 = sco=co n-3 (5 6-3.2 5. 4 n = 4, с с 6 - - 4.3 6.5 •Cal 65 sa= – V5.co n-5, са 6- 5.4 7.6 تی) - 16 - 17.0=0 mak, C8 6- Bor 6.5 oc6 -24 . 1-Ys Co) 3 335 y = cotcom +3 content controns-coat ton7 35 Cox 8 since &(o)- 3 an G'(o)=3 C-3 Co 3 and

we get y (2)= 1 3nen4 - Yn6ut 3₂5284 y = coll+an+n" - Ysnb & 2 5 8 8 .-) ec.(x). comparing this with Y(n) comparing His @ and Yon) = Cine = 30. 3. [:c1=3 with we get, Hien le Ž annan = coGiCnt (2 G2 (2) 3₂5 a = 3 az=1, a3=-/s, a 4 = Finally the polynomial Y₂ (2)=3R.