Solving a Recurrence Relation with Characteristic Roots: is my algebra bad, or is it something else? I am having a considerable amount of trouble solving this problem. I have shown my work below and am wondering if somebody can show me where I have messed up. I have been working on this for a few hours and would appreciate the help.
The question

Solve $a_0$ $=$ $1$, $a_1$ $=$ $3$, and $a_{n-1}$ +$6a_{n-2}$ for $n$ $\geq$ $2$.


My Work
$$a_n-a_{n-1}-6a_{n-2}=0$$
$$r^{2}-r-6=0$$
$$(r-3)(r+2)$$
$$r=3, r=-2$$
then, $$\alpha(3)^{n}+\beta(-2)^{n}=a_n$$
and 
$$a_0=\alpha(3)^{0}+\beta(-2)^{0}=1$$
thus,$$\alpha +\beta=1$$
and $$a_1=\alpha(3)^{1}+\beta(-2)^{1}=3$$
By simplifying, $$-3(\alpha+\beta=1)$$
$$3\alpha-2\beta=3$$
$$\beta=0, \alpha=1$$

I know this is not the case. I have tried switching around the $-2$ and $3$ in the equation, but I can't seem to get the answer that way either. I end up with coefficients equalling $-\frac{1}{5}$ and $\frac{4}{5}$ or something to that effect. I am struggling on this problem and would appreciate guidance. 
 A: To summarize the many comments under the main question, OP's solution is correct, and $\,a_n=3^n\,$ is indeed the correct result. Step by step...

The question
Solve $a_0$ $=$ $1$, $a_1$ $=$ $3$, and $a_{n-1}$ +$6a_{n-2}$ for $n$ $\geq$ $2$.

Presumably, the latter part should read $\,\color{red}{a_n=}a_{n-1} + 6a_{n-2}\,$ for the recurrence to be well defined.

My Work
$$a_n-a_{n-1}-6a_{n-2}=0$$
$$r^{2}-r-6=0$$
$$(r-3)(r+2)$$
$$r=3, r=-2$$

Correct. The given relation is a linear homogeneous recurrence with constant coefficients, and one standard method to solve it is using the associated characteristic equation.

$$\alpha(3)^{n}+\beta(-2)^{n}=a_n$$

Correct, $\,a_n=\alpha \cdot 3^n + \beta \cdot (-2)^n$ where $\,\alpha,\beta\,$ are constants to be determined from the initial conditions.

$$\beta=0, \alpha=1$$

Correct. Substituting back, it follows that $\,a_n=\alpha \cdot 3^n + \beta \cdot (-2)^n = 1 \cdot 3^n + 0 \cdot (-2)^n = 3^n\,$.

I know this is not the case.

This in fact the only wrong statement of the post. It is the case that $\,a_n=3^n\,$ is the correct solution. For verification:

*

*$a_0=3^0=1\,$ satisfies the first initial condition;


*$a_1=3^1=3\,$ satisfies the second initial condition;


*$a_n=3^n=3 \cdot 3^{n-1} = 3^{n-1} + 2 \cdot 3^{n-1} = 3^{n-1} + 2 \cdot 3 \cdot 3^{n-2} = a_{n-1}+6 a_{n-2}$ satisfies the recurrence relation for $\,n \ge 2\,$.
