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I have a second-order difference equation which satisfies the following:

$\frac{a_{n+2}-a_{n+1}}{a_{n+1}-a_{n}} = k >0$ for any $n$.

I can see that $ \Delta a_{n}\equiv a_{n+1}-a_{n}$ is satisfying an exponential relation, but what more can I say about $a_{n}$ itself?

Thank you very much

Anonymously lost student
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  • The relation can be rewritten in the form $a_{n+2} = (k+1)a_{n+1} - k a_n$ which shows that given $a_0, a_1$, every successive $a_n$ for $n \geqslant 2$ is determined. Then the standard way of solving it is then to try $a_n = \alpha x^n$ and obtain possible values of $\alpha$. – WA Don May 20 '21 at 14:06
  • Thanks for the answer! I came up with that exponential solution, but it didn’t seem to be the unique solution. Is that correct? For instant $a_{n}=\alpha k^{n} + \beta k^{n-1}$ would also satisfy.. – Anonymously lost student May 20 '21 at 14:33
  • If $a_0$ and $a_1$ are given there will be just one solution. If these are not specified the set of solutions will be fully described by linear combinations of $k^n$ and a constant (i.e. $a_n = \alpha k^n + \beta$) – WA Don May 20 '21 at 15:41

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