Question

2a and 2b!! im not sure how to plug in / do the Initial values??

Problem 1. Find the general solution to each of the following homogeneous difference equations: a) It+3+ 5x+2 - It+1 - 524 = 0 c) 14+3 - 314+2 +It+1 -3.14 = 0 b) 14+4 - 8x++2 +16x = 0. d) 1614+4 - 8x2+2 + It = 0. Problem 2. For equation a) from Problem 1, find the unique solution satisfying to = I1 = 0 and I2 = 5. Also, for equation b) from Problem 1, find the unique solution satisfying Xo = x2 = 0 and 11 = I3 = 2

Answer 1

a team za Let it to the fesward and shifts the index shift operater; that takes are step forward to Death le. En = Katl Ex = E(Ex) = Elanee) = Knez Erna khari =) solve x3 + 5742 7441 - 57450 The above equation can be written in the form (5€-E-5134 so And the rests of E+SE²_E-520 The moltes of the above squ are E=1,-1,-5 The general suht is 4.= acist & brist - cent 14 = at beitt cerst 7 0 Giver initial condition are to=x, 20, and 22 = 5 pt 26=0 = atblot ceri. : 0 = ätbtc => [aebte zo]-> pit 2:4 in o we get o= at bli't cft) . a-b-50.20 + put zas in o he get 5 = atblent care =>: 5:= a+b+850 .: ' fat b+ 256 = + solving &.. @ @ we get a = sa ba-se ca 5 ca The site is 4 = 5 - Fast to see a { 10 + 15+15+ tsent] in © we get

2) 444-844+2 +16 24 =0 with x=7920 x = 22=2 - The above can can be written as (E9_862 +16)* 30 we heas have to solve 64-8E+16=0 . (-4)2=0 €.4,4 ... The roots are 2 2 -2 -2 " . - The general son is x = Q.bt) fait + (Cedt) cast giver that zo zo The 6» (a +014-2) + C+ d.o) 2 = 0 i a c =0 x=g then 0 - (a+b) (-2) + ((td) 2' =2 → - x=0 then 6 ) - X₂=2 than » -20-2b + 2C tad = 2 -acob +c+d = 1 Q+Qb) fait 642d) 2 = 0 40+8b+ 4C+8d = 0 arab+QC + ad =0 tit3b) tasid + (+34) 2 = 2 8( at 3b) + 8C(+3d) = 2 -40-18b +4C18d = 1 upan solving equatias @.6 @ and we get a=h to be calho d= -3/16 The study is 7 = 6***+) 6234 + 64 - #+) at 1 t[ 41143+(-2+ + (1 - st)[ty] ->