$a_n=5a_{n-1} - 6_{n-2} + 4^n + 2n + 3$ for $n>=2$ , $a0 = 5, a1 = 19.$
I get the general solution
$ c_n = C_12^n+C_23^n.$
For a particular solution in the form $pn = An + B + C4^n$; we have
$An + B + C4^n = 5(A(n − 1) + B + C4n−1) − 6(A(n − 2) + B + C4n−2) + 4n + 2n + 3$
But how do we get this:
$An + B + C4^n = −An + 7A − B + 14C4^{n−2} + 4^n + 2n + 3$
and therefore this:
$2An − 7A + 2B + 2C4^{n−2} = 2n + 3 + 16 × 4^{n−2}$
I don't understand it.
Hint:
Use the *principle of superposition of solutions; find particular solutions of $$a_n=5a_{n-1} - 6_{n-2} + 4^n\quad (\text{resp. }{} + 2n, {}+ 3)$$ then add these particular solutions.
Example for $4^n$:
A particular solution will have the form $a_n=C 4^n$. It is indeed a solution if $$C4^n=5C4^{n-1}-6C4^{n-2}+4^n\iff 16C=20C-6C+16\iff C=8.$$