Question

Consider the following differential equation. (1 + 6x2)y" – 4xy' – 24y = 0 (a) If you were to look for a power series solution about xo = 0, i.e., of the form onth n=0 then the recurrence formula for the coefficients would be given by ck+2 = g(k) Ck, k > 2. Enter the function g(k) into the answer box below. (b) Find the solution to the above differential equation with initial conditions y(0) = 0 and y'(0) = 5. (Note that this solution is a terminating power series.) (c) Find the first three nonzero terms in the solution to the above differential equation with initial conditions y(0) = 3 and y'(0) = 0.

Answer 1

(1+682) ( in - 19 con 2*-2 ) - 4* (Iron 2) - 24 (cnx"): 0 Inm-1) Cn x n -2 + 6 En.n-1) cnxh - 4žnnur - 24 Z cox - 20 00 ne2 y' = I nnn- ; " = n(n-1) on 1-2 (60+8)(n-3) cn. (1+68²) y" - quy' – 24 y = 0 lel, y = Zona" no ni na 2 6 Then, 43 non n=2 - 2 nao n=2 h2 ni I (n+1)(n+2) Gn+2 + n1 no n = 2 + 6 In(n-1) cn x² 202 + 6C3 X + I (n+1)(n+ 2) Cn +2 n2 n=2 -4c, a 4 nanan - 24 Co -24 GX – 24 I Cna" 2. (20₂-2460)+(6C3 - 401-246)x+ I ((n +1)(n+2) Cn+2 +6n(n-1) cn - anen - 246]>" = - 0 (2C2 - 24 60 ) + (6(3 - 2801 ) 2 + +(6n²_ion-24) an an] aan = E [(n+1) (n +2) Cn +2 2c2-24co 7 C2 = 12co 6c3-284 is C3 = 140 (n+1) (n + 2) (n + 2 + C6n²_lon-24) n = 0 6n? - ion-24 6n²-18nton-24 an- Cn+ 2 = (n+1) (n + 2) (n+1) (n + 2) on(n-3) +8 (n-3) Cn - (n+1)(n+2) (n+1) In +2) n=2 n2 U n=2 رنى =0 & Cn so, CK+2 = Ск - CV (6K+8) (K-3) (K+1) (K+2) = g(K) CK Ck. Where, 9(K) (6K+8)(K-3) (K+1) (K+2) b) now, ycol = 0, y 10) = 5 y'(o) = 5 SO Ad Co = y0) = 0; C = - from i C2 = 12 = 0 C3 - 14 xs Ho 3 form i = 16G Ruch 8 put k=2 = 0. C2 C4 20.(-1) 3.4 SAXO k=3 26. (O) cs -0. C3 4.5 K=4 32. 1 CC cu - o 56 K=s C7 Il 38.2 6.7 Therefore CK V Ky, 7 since, CS=0=4=0 0720

Therefore, y = ²axh no = Cotak + C₂ x ² + 233 = 5x + 70 x c) yeo) = 3, y'co) = 0 so, Co = 3 so, C2 = 12 Co = 36 c3 = 14 Cu 20.(-1) 3 u C2 = 20.(-1) 3.4 X 36 = 60 C: C3 =0) C6 32:1 ll Cu 3 2 x60 =-64 3o Therefore, + y = 3 +36x² + 60x4_640 26