Suppose I have the following difference equation \begin{equation} v(t+2) + a v(t+1) + b v(t) = 0. \end{equation} I wish to convert it to a second order ODE. How does one go about doing this? (note: initial conditions are arbitrary).
I have tried using a taylor series where I introduced $v(t+2) \to v(t+2h)$ and so on, but this did not lead anywhere.
Any help would be appreciated
Note that $2\left( {v(t+2) - v(t)\over 2} \right )+ a\left({v(t+1) - v(t) \over 1}\right) + (b + a + 1)v(t) = 0$. Convert this to
$$2\left( {v(t+2h) - v(t)\over 2h} \right )+ a\left({v(t+h) - v(t) \over h}\right) + (b + a + 1)v(t) = 0$$
and let $h \to 0$ to get $(2 + a)\dot v + (b + a + 1)v = 0$.
That is one possible way to convert it to an ODE. There are others. You'll note that it isn't particularly useful unless $ t \gg 1$.
Edit: To make this a 2nd order ODE,
If $v'(t) \leftrightarrow v(t+1) - v(t)$, then $$\begin{align}v''(t) &\leftrightarrow v'(t+1) - v'(t) \\&\leftrightarrow (v(t+2) - v(t+1)) - (v(t+1) - v(t)) \\&\leftrightarrow v(t+2) + v(t) - 2v(t+1).\end{align}$$
So we rearrange the original expression to $$(v(t+2) + v(t) - 2v(t+1)) + (a +2)(v(t+1) - v(t)) + (b + a + 1)v(t)$$ and convert to $$v'' + (a+2)v' +(b + a + 1)v = 0.$$