Amateur Math and a Linear Recurrence Relation I haven't received a formal education on this topic but a little googling told me this is what I am trying to find. I would like to put
$$ a_n = 6 a_{n-1} - a_{n-2} $$
$$ a_1 =1, a_2 = 6 $$
into its explicit form. So far I have only confused myself with billions of 6's.
I would like someone to respond with the following:


*

*Is this relation homogenous? (not sure if subtraction counts)

*Is there a basic way I can solve this type of recurrence? (I don't need the long answer, just enough to satisfy curiosity)
I am not necessarily looking for only a solution to this problem but one would be appreciated nonetheless. Still, I am more interested in the background and would better appreciate that than simply some formula.

As user164587 answered, the closed form is $$a_n = A\alpha_1^n + B\alpha_2^n$$ where $\alpha_1$ and $\alpha_2$ are the roots of the characteristic equation $u^2 - 6u + 1 = 0$ that is to say $\alpha_1=3-2 \sqrt{2}$, $\alpha_2=3-2 \sqrt{2}$. 
Solving for $A$ and $B$, you would find that $$A=-\frac{{a_2}-{a_1} {\alpha_2}}{{\alpha_1} ({\alpha_2}-{\alpha_1})}$$ $$B=-\frac{{-a_2}+{a_1} {\alpha_1}}{{\alpha_2} ({\alpha_2}-{\alpha_1})}$$
Using the values of $a_1,a_2,\alpha_1,\alpha_2$, you then arrive to $$a_n= \frac{\left(3+2 \sqrt{2}\right)^n-\left(3-2 \sqrt{2}\right)^n}{4 \sqrt{2}}$$ which is a simple expression. 
