# About an non-autonomous difference equation

Let $r>4$ be a positive integer. Let us consider this difference equation: $$v_{n+1}-r^{2n+1}v_{n}=w$$ where $w$ is real number.

How I can solve this equation with respect to $v_{n}$. I am not able to find the good idea.

• This is basically an interest problem with deposits and a growing interest rate. I do not know a closed for for such problems however. – Ian Oct 2 '15 at 10:05
• @Ian: Did you mean that it is impossible to solve this equation. – DER Oct 2 '15 at 10:15
• No, I just do not know whether a closed form exists. If you unfold a little you will see it looks a bit like a geometric series, so maybe you can exploit that. – Ian Oct 2 '15 at 10:19

## 2 Answers

Let's find an "integrating factor" where the quotient is $r^{2n+1}$. As $(n+1)^2-n^2=2n+1$, we get that $$\frac{v_{n+1}}{r^{(n+1)^2}}-\frac{v_n}{r^{n^2}}=\frac{w}{r^{(n+1)^2}}$$ The summation over the square-degree powers of $r^{-1}$ however has no elementary simplification, so that $$\frac{v_n}{r^{n^2}}=\frac{w}{r^{n^2}}+\frac{w}{r^{(n-1)^2}}+…+\frac{w}{r^{1}}+\frac{v_0}{r^0}$$ is as simple as it gets.

We have $v_{n+1}=w + r^{2n+1} v_n$, hence $$v_{n+2} = w + r^{2n+3} v_{n+1} = w + r^{2n+3} w + r^{4n+4} v_n$$ and by induction:

$$v_{n+k} = w\sum_{j=1}^{k}r^{2jn+k^2-(k+1-j)^2}+v_n r^{2kn+k^2}.$$