Question

(1 point) In this exercise we consider the second order linear equation y" + series solution in the form y = 0. This equation has an ordinary point at x = 0 and therefore has a power y = cmx". n=0 We learned how to easily solve problems like this in several different ways but here we want to consider the power series method. (1) Insert the formal power series into the differential equation and derive the recurrence relation Cn = Cn-2 for n = 2,3,... NOTE Co and ci are arbitrary constants. The recurrence relation then implies that all the even terms C2k for k = 1,2, ..., are given in terms of co and all odd terms C2k+1 for k = 1, 2, , are given in terms of c1. Thus we see that the solution series can be written as a sum of two series consisting of the even and odd parts of the original series. Therefore the general solution can be written as y(x) = Ayı (x) + Byz(x) where A = Co and B = ci are arbitrary constants and =1+302x2*, »=x+ au 21 (2) Use the recurrence relation to find the first few coefficients in terms azk and a2k+1: 22 = ,24 = .06 = az = a5 = , az = (3) Finding the general patterns is a bit more challenging so we will simply write y = [1+ x2+ ** + ---), yz = [x+ x+ x + ...] Notice The general solution of this first order linear equation is y = co cos(3x) + C sin(3x)

Answer 1

2 de M=0 1-2 I, nlnyc x n=2 y+9y=0 > 0 Let y = 2 2 4 2h >>g's Simenosimo - O+ 8 n 1)(x * 3 8 GB PE (2+2)(n+1) Short + $.40670 . [(na) (n =a/indet 9 cu]a!-o (11- n-2 XC HBO 1-2 -> (1+) (2+2) Catet 9c, zo, th), =) Entz = -9 cho . (1+1)(n+2) 3 422, C₂ = -9 26-02-20 133, 5z = =39 1=4, C4 = 72/2 C2 = 27 co 1-5, C5 = 54 53 = 24 (13) n=6, C6= -26 4 = - loco n=1, C4 = 22 16 5 = el loro () = (1 - 1 1 ...) 5 1. X - 81, 560