Consider the following problem:
Consider the sequence defined by $a_n=xa_{n-1}+ya_{n-2}$. Find the general formula for $a_n$ in terms of $a_1$ and $a_2$.
This is a pretty typical recursion. My question is: is there a way to solve this recursion without using the characteristic polynomial?
As an example, consider the following problem: (sorry for my excessive use of "consider")
Consider the sequence defined by $a_n=xa_{n-1}+y$. Find the general formula for $a_n$ in terms of $a_1$.
A pretty elegant way of doing this is to write:
$$ a_n=xa_{n-1}+y \\ a_n+\frac{y}{x-1}=xa_{n-1}+y+\frac{y}{x-1}=xa_{n-1}+\frac{yx}{x-1} \\ \begin{align} a_n+\frac{y}{x-1} &=x(a_{n-1}+\frac{y}{x-1}) \\ &=x^2(a_{n-2}+\frac{y}{x-1}) \\ &=x^{n-1}(a_1+\frac{y}{x-1}) \end{align} $$ The problem can now be easily solved from here.
Another example is the following problem:
Consider the sequence defined by $a_n=x_na_{n-1}+y_n$ where $\{x_i\}$ and $\{y_i\}$ are also both sequences. Find the general formula for $a_n$ in terms of $a_1,x_i$ and $y_i$.
The solution can be found here.
I have browsed many questions and articles on recursions, like this very well explained article, which unfortunately does not answer my question, and this question. The first answer uses computer to find a pattern before proving that the pattern works, but in the context of Mathematics Olympiads, this method is clearly not viable. I could not understand the second answer, and the third answer uses generating functions. However, I am looking for an elementary and nice, elegant method of doing this, like I have shown above.
So, is there such a method? If yes, can this be generalised to recursions like $a_n=xa_{n-1}+ya_{n-2}$ or $a_n=xa_{n-1}^2+y$? Any help will be appreciated and, at my discretion, awarded with bounties.
