Could anyone assist me in solving the following recurrence relations?
$a_n = 3a_{n-1} - 2a_{n-2} + 2^n n^2$
$b_n = -nb_{n-1} + n!$
Specifically, I am not sure how to find the particular solutions to such relations once a homogeneous solution is given?
On
For the first question, the techniques that Wilf's "generatingfunctionology" present in the second chapter make the solving of such routine.
The last part of this answer tells you how to deal with the first problem.
For the second, we need initial conditions. If $b_0=0$, however, you can do it almost by inspection: taking $b_0=0$ generates the sequence $\langle 0,1!,0,3!,0,5!,\dots\rangle$, and it’s not hard to guess and then show by induction that
$$b_n=\begin{cases}0,&\text{if }n\text{ is even}\\n!,&\text{if }n\text{ is odd}\end{cases}$$
is a solution. In fact, you can extend that idea to get a general solution:
$$\begin{align*} &b_0=b_0\\ &b_1=-b_0+1!\\ &b_2=2b_0-2!+2!=2!b_0\\ &b_3=-3!b_0+3!\\ &b_4=4!b_0-4!+4!=4!b_0\;, \end{align*}$$
and you can guess and prove by induction that in general
$$b_n=\begin{cases} n!b_0,&\text{if }n\text{ is even}\\ -n!b_0+n!,&\text{if }n\text{ is odd}\;. \end{cases}$$
Added: It just occurred to me that we could also change the variable. Let $c_n=\frac{b_n}{n!}$; then after division by $n!$ the recurrence $b_n=-nb_{n-1}+n!$ can be written $c_n=-c_{n-1}+1$. This recurrence can easily be solved by any number of techniques, and then it’s just a matter of restoring $b_n$ by multiplying by $n!$.