Solve the non-homogeneous recurrence relation $\ a_n = 12 a_{n−2} − 16 a_{n−3} + 9 · 2^ {n+1}+ 25 n, n ≥ 3,\ where\ a_0 = 20, a_1 = 31 , a_2 = −62$

Question

$\ a_n = 12 a_{n−2} − 16 a_{n−3} + 9 · 2^ {n+1}+ 25 n, n ≥ 3,\ where\ a_0 = 20, a_1 = 31 , a_2 = −62$

How do I solve this? I understand the steps but I'm stuck for the particular solution of the non-homogeneous part and the general solution of the homogenous part... Also, does the$\ 9 . 2^{n+1}$ part go to both the non-homo & homo part?

Thanks!!

Answer 1

Denote $a_n = b_n + c_n $ where

$b_n = 12 b_{n−2} − 16 b_{n−3} $
$c_n = 12 c_{n−2} − 16 c_{n−3} + 9 · 2^ {n+1}+ 25 n$
$n ≥ 3, $
$b_0 = c_0 = 20, b_1 = c_1 = 31 , b_2 = c_2 = −62 $

Solving the homogenous part usual way and the inhomogenous part by "enlightened quess",
( for example, $ c_n = K_1n2^n + K_2n^2 + K_3n$ might work ) and combining the answers gives the result for the original equation $a_n$.