We will find the solution to the following lhcc recurrence:
$a_{n} = 2 a_{n-1} + 1 a_{n-2} - 2 a_{n-3} \text{ for } n \geq 3$ with initial conditions $a_0 = 13, a_1 = 5, a_2 = 19$. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format $a_n = r^n$. (We assume also $r \neq 0$). In this case we get: $$r^n = 2 r^{n-1} + 1 r^{n-2} - 2 r^{n-3}.$$ Since we are assuming $r \neq 0$ we can divide by the smallest power of r, i.e., $r^{n-3}$ to get the characteristic equation: $$r^3 = 2 r^2 + 1 r - 2.$$ (Notice since our lhcc recurrence was degree 3, the characteristic equation is degree 3.)
Find the three roots of the characteristic equation $r_1,r_2$ and $r_3$. When entering your answers use $r_1 \leq r_2 \leq r_3$:
$r_1 =$ , $r_2 =$ , $r_3 =$ .

Since the roots are distinct, the general theory (Theorem 3 in section 5.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: $$a_n = \alpha_1 (r_1)^n + \alpha_2 (r_2)^n + \alpha_3 (r_3)^n$$ for suitable constants $\alpha_1, \alpha_2, \alpha_3$.
To find the values of these constants we have to use the initial conditions $a_0 = 13, a_1 = 5, a_2 = 19$. These yield by using n=0,n=1 and n=2 in the formula above: $$13 = \alpha_1 (r_1)^0 + \alpha_2 (r_2)^0 + \alpha_3 (r_3)^0$$ and $$5 = \alpha_1 (r_1)^1 + \alpha_2 (r_2)^1 + \alpha_3 (r_3)^1$$ and $$19 = \alpha_1 (r_1)^2 + \alpha_2 (r_2)^2 + \alpha_3 (r_3)^2$$ By plugging in your previously found numerical values for $r_1, r_2$ and $r_3$ and doing some algebra, find $\alpha_1, \alpha_2, \alpha_3$:
Note: Ad hoc substitution should work to find the $\alpha_i$ but for those who know linear algebra, note the system of equations above can be written in matrix form as: $$\begin{bmatrix} (r_1)^0 & (r_2)^0 & (r_3)^0 \\ (r_1)^1 & (r_2)^1 & (r_3)^1 \\ (r_1)^2 & (r_2)^2 & (r_3)^2 \end{bmatrix} \begin{bmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \\ 19 \end{bmatrix}$$ $\alpha_1 =$
$\alpha_2 =$
$\alpha_3 =$

Note the final solution of the recurrence is: $$a_n = \alpha_1 (r_1)^n + \alpha_2 (r_2)^n + \alpha_3 (r_3)^n$$ where the numbers $r_i, \alpha_i$ have been found by your work. This gives an explicit numerical formula in terms of n for the $a_n$.