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I'm currently being introduced to $2^{nd}$ order linear homogenous recurrence relations for the first time. I was working through a first example in my textbook and came into some trouble. Here is the section of my textbook:

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What is confusing me is how they write $a_n = c_1(2^n) + c_2(-3)^n$ as the general solution near the end of the second paragraph. Why do we add our two solutions when it is clear $a_n = 6a_{n-2} - a_{n-1}$ from what we were given in the first line?

Dunka
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We want a formula for $a_n$ that doesn't involve other $a$'s, and thus allows us (once we know the constants $c_1$ and $c_2$ to calculate any $a_n$ just from $n$.

Robert Israel
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  • Hmmm, I think I understand. It's legal because we are just saying there are two constants we can put in front of our obtained values of $a_n$ such that their sum(or difference in the case of a negative constant) would be equal to $a_n$? – Dunka Feb 24 '15 at 05:08
  • It's legal because a linear combination of solutions of a linear equation is a solution. – Robert Israel Feb 24 '15 at 05:13
  • Okay, but isn't the equation basically saying $a_n = c_1a_n + c_2an$ meaning we're writing it in such a way that we know there are two constants whose product with $a_n$ will sum to $a_n$? – Dunka Feb 24 '15 at 05:17
  • No, that's not true at all. You have two different solutions ($2^n$ and $(-3)^n$), and every solution is some constant $c_1$ times the first plus some constant $c_2$ times the second. – Robert Israel Feb 24 '15 at 16:10
  • Aren't the two solutions on their own? I'm definitely a little lost here. – Dunka Feb 24 '15 at 20:01
  • What do you mean by "on their own"? I don't understand what's bothering you about this. – Robert Israel Feb 24 '15 at 22:41