Solve the following recurrence relation:
$$a_n = 3a_{n-2}+2a_{n-3} + 81n^2(2)^n+32(3)^n+4n+4$$
Workings:
$a_n^{(h)} = 3a_{n-2}^{(h)}+2a_{n-3}^{(h)}$
$ch(x) = x^3 + 3x^2 + 2x$
$ch(x) = x(x+1)(x+2)$
$(1) a_n = 3a_{n-2} + 2a_{n-3} + 81n^2(2)^n$
$(2) a_n = 3a_{n-2} + 2a_{n-3} + 32(3)^n$
$(3) a_n = 3a_{n-2} + 2a_{n-3} + 4n + 4$
$(1) a_n^{(p_1)} = (c_1n^2 + c_2n +c)(2)^n$
$(c_1n^2 + c_2n +c)(2)^n = 3(c_1(n^2-2)+c_2(n-2)+c_3)2^{n-2} + 2(c_1(n^2-3)+c_2(n-3)+c_3)2^{n-3} + 81n^22^n$
$8(c_1n^2+c_2n+c_3) = 6(c_1(n^2-2)+c_2(n-2)+c_3) + 2(c_1(n^2-3)+c_2(n-3)+c_3) + 648n^2$
$n^2: 8c_1 = 6c_1 + 2c_1 + 648$
$n: 8c_2 = 6c_1 + 2c_2$
$1: 8c_3 = -12c_1 -12c_2 + 6c_3 - 6c_1 - 6c_2 + 2c_3$
Now I'm stuck and don't sure what to do. Any help will be appreciated.
This might not be explicitly solvable, due to the unsolvability of quintics. But a general approach is "write the linear parts on one side of the equation, and the non linear parts on the other, then increment":
$$\begin{align} % a_{n} - 3 ~ a_{n - 2} - 2 ~ a_{n - 3} &= 81 ~ n^2 ~ 2^n + 32 ~ 3^n + 4 ~ n + 4 \\ % a_{n + 1} - 3 ~ a_{n - 1} - 2 ~ a_{n - 2} &= 162 ~ n^2 ~ 2^n + 324 ~ n ~ 2^n + 162 ~ 2^n + 96 ~ 3^n + 4 ~ n + 8 \\ % a_{n + 2} - 3 ~ a_{n} - 2 ~ a_{n - 1} &= 324 ~ n^2 ~ 2^n + 1296 ~ n ~ 2^n + 1296 ~ 2^n + 288 ~ 3^n + 4 ~ n + 12 \\ % a_{n + 3} - 3 ~ a_{n + 1} - 2 ~ a_{n} &= 648 ~ n^2 ~ 2^n + 3888 ~ n ~ 2^n + 5832 ~ 2^n + 864 ~ 3^n + 4 ~ n + 16 \\ % a_{n + 4} - 3 ~ a_{n + 2} - 2 ~ a_{n + 1} &= 1296 ~ n^2 ~ 2^n + 10368 ~ n ~ 2^n + 20736 ~ 2^n + 2592 ~ 3^n + 4 ~ n + 20 \\ % a_{n + 5} - 3 ~ a_{n + 3} - 2 ~ a_{n + 2} &= 2592 ~ n^2 ~ 2^n + 25920 ~ n ~ 2^n + 64800 ~ 2^n + 7776 ~ 3^n + 4 ~ n + 24 \\ % a_{n + 6} - 3 ~ a_{n + 4} - 2 ~ a_{n + 3} &= 5184 ~ n^2 ~ 2^n + 62208 ~ n ~ 2^n + 186624 ~ 2^n + 23328 ~ 3^n + 4 ~ n + 28 \\ % \end{align}$$
Now the nonlinear side of the equations are composed of six coefficients:
$$\begin{cases} n^2~2^n = u \\ n~2^n = v \\ 2^n = w \\ 3^n = x \\ n = y \\ 1 = z \\ \end{cases}$$
$$\begin{align} % a_{n} - 3 ~ a_{n - 2} - 2 ~ a_{n - 3} &= 81 ~ u + 32 ~ x + 4 ~ y + 4 ~ z \\ % a_{n + 1} - 3 ~ a_{n - 1} - 2 ~ a_{n - 2} &= 162 ~ u + 324 ~ v + 162 ~ w + 96 ~ x + 4 ~ y + 8 ~ z \\ % a_{n + 2} - 3 ~ a_{n} - 2 ~ a_{n - 1} &= 324 ~ u + 1296 ~ v + 1296 ~ w + 288 ~ x + 4 ~ y + 12 ~ z \\ % a_{n + 3} - 3 ~ a_{n + 1} - 2 ~ a_{n} &= 648 ~ u + 3888 ~ v + 5832 ~ w + 864 ~ x + 4 ~ y + 16 ~ z \\ % a_{n + 4} - 3 ~ a_{n + 2} - 2 ~ a_{n + 1} &= 1296 ~ u + 10368 ~ v + 20736 ~ w + 2592 ~ x + 4 ~ y + 20 ~ z \\ % a_{n + 5} - 3 ~ a_{n + 3} - 2 ~ a_{n + 2} &= 2592 ~ u + 25920 ~ v + 64800 ~ w + 7776 ~ x + 4 ~ y + 24 ~ z \\ % a_{n + 6} - 3 ~ a_{n + 4} - 2 ~ a_{n + 3} &= 5184 ~ u + 62208 ~ v + 186624 ~ w + 23328 ~ x + 4 ~ y + 28 ~ z \\ \end{align}$$
Now the 6 variables can be canceled from the 7 linear equations, to form 1 equation:
$$a_{n + 6} - 11 ~ a_{n + 5} + 46 ~ a_{n + 4} - 82 ~ a_{n + 3} + 17 ~ a_{n + 2} + 149 ~ a_{n + 1} - 176 ~ a_{n} - 8 ~ a_{n - 1} + 112 ~ a_{n - 2} - 48 ~ a_{n - 3} = 0$$
Which can be written in matrix form as:
$$ \begin{bmatrix} a_{n+9} \\ a_{n+8} \\ a_{n+7} \\ a_{n+6} \\ a_{n+5} \\ a_{n+4} \\ a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ a_{n} \end{bmatrix} = \begin{bmatrix} 1 & -11 & 46 & -82 & 17 & 149 & -176 & -8 & 112 & -48 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix}^n \begin{bmatrix} a_{9} \\ a_{8} \\ a_{7} \\ a_{6} \\ a_{5} \\ a_{4} \\ a_{3} \\ a_{2} \\ a_{1} \\ a_{0} \end{bmatrix}$$