Question

Please write neat so I can read and understand the problem.

(1 point) In this exercise we consider finding the first five coefficients in the series solution of the first order linear initial value problem 3y" - xy + 4y = 0 subject to the initial condition y(0) = 3, y'(0) = 2. Since the equation has an ordinary point at x = 0 and it has a power series solution in the form y= 2" We learned how to easily solve problems like this separation of variables 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 2 for n = 2,3,... The solution to this initial value problem can be written in the form y(c) = Coyi (2) + C 422) where co and G are determined from the initial conditions. The function yı (2) is an even function and y2(x) is an odd function. For this example, from the initial conditions, we have co = and q = . o The function y2() is an infinite series y22) = x + a2k1122k+1. k=1 (2) Use the recurrence relation to find the first few coefficients of the infinite series Az = .a5 = an= NOTE note that the constant c has been factored out. NO TE In the function yı (2) the first term is 1. In The function yı (2) is an even degree polynomial yı = other words, note that the constant Co has been factored out.

Answer 1

no 34" - ny' +4y=0 y lo)=3, y: 10)-2 let solution is y. & Conx? Now, y'= Encours y" (P-1) Co 29-12 - 3 D (MM) (m 20-2En Cox' + 4 cm = 0 Cont nzo n-2 no

=> 3 en (n-1) Comento Core?+u CM 22 9 = 0 2 Now index in last two terr tror. o an Shifting to 2-2 0 - 9 1 =) 3 & n(n-1) Con- non -2 +4E (n-22 =0 n =22 ne2 n-3 ->[3(2)(0)62 +4C0722° + ¿ 613 (0 -) -10-3)(n-2+46->) n 22. equating to loetficient of zero no 66a+uco=0 (12--2 10 / Reccurence orekation is 3 (0-1) Cn - (n-2) (-28 + 460-2-0 2) 30 (0-1) (n = Cn-a (4+0-2) n = n=2,3, (n+2) Cn-2 3n (nt)

» 3 - соу" - Со + Cl2 + (ә я? - 3 = 910) = Со т' (1) -- C, + 2 Can + - - 2 - 'to) - с. » (Coe 3, С-г / Сл- (n+2 ) , Co-a Со е 2, С3- 5 xc, - 5 3х3х8 -5 Сча 6 . Cas k x (2 (3) ЗX & 2 1 7 cs : sХ Ч

30»); ; com. — zo - Со + Со»?, с » ч. 4, 6, 7 * Су » Са»", Cs * Су”, - 7 8. (m) е Cot tan® + Cy +9+ C6 6 + - Со [++ (-2) »?», 7 ON 9а (7) , C, 1 + C3 2 + CsS + C27+ + 7 - S .4 + 7, сі и за - «[* * * * ”, - 7 23 5,45 - 1, 4- у