This is the recurrence relation:
$a_n=5a_{n-1} - 6a_{n-2} + 4^n + 2n + 3$ for $n\geq2$ , $a_0 = 5, a_1 = 19.$
I get the general solution.
$ c_n = C_12^n+C_23^n.$
The particular solution is in the form
$p_n = An + B + C4^n$
Can someone explain in detail how I actually find the particular solution with a solution themselves?
Any help is appreciated
$$\begin{array} {rl} % a_{n} &= 5a_{n-1} - 6a_{n-2} + 4^n + 2n + 3 \\ % a_{n+1} &= 5a_{n} - 6a_{n-1} + 4^{n+1} + 2(n+1) + 3 \\ % a_{n+2} &= 5a_{n+1} - 6a_{n} + 4^{n+2} + 2(n+2) + 3 \\ % a_{n+3} &= 5a_{n+2} - 6a_{n+1} + 4^{n+3} + 2(n+3) + 3 \\ % \end{array}$$
Then let $x = 4^n$, $y = 2n$, $z = 1$, plug that into the above 4 equations to get 4 linear equations that are linear in $a_{*}$ and $x, y, z$. Eliminate $x,y,z$ from the equations to get 1 homogenous equation, note that the roots of the original equation $a_n = 5a_{n-1} - 6a_{n-2}$ will also be roots of the new homogenous equation.