Discrete time equivalent to ODE I'm reading a paper in which it is noted that
$$\frac{dv(t)}{dt} = f(t) - \varepsilon v(t)$$
has the discrete time equivalent
$$v(t+1) = v(t)\exp(-\varepsilon) + \frac{f(t)}{\varepsilon}[1 - \exp(-\varepsilon)].$$
How is this derived?
Update:
In case I have misunderstood, my question is a simplified version of the text appearing at the top of Page 2 in this supplementary material from this paper.  Namely
$$\frac{dv_i}{dt}=\frac{I_i}{N_i} - \varepsilon v_i$$
$$v_i(t+1)=v_i(t)\exp(-\varepsilon)+\frac{I_i}{\varepsilon N_i}[1-\exp(-\varepsilon)]$$
Where I understand $\tfrac{I_i}{N_i}$ to be the evolving proportion of infectious animals at a location.
 A: Subtracting $ v(t) $ from either side gives:
$$
v(t+1) - v(t) = v(t) \exp(-\epsilon) - v(t) + \frac{f(t)}{\epsilon}( 1 - \exp(- \epsilon ) )
$$
As $ \epsilon \rightarrow 0 $, taylor expanding the exponentials, $ \exp(-\epsilon) = 1 - \epsilon + \epsilon^2 / 2 \cdots $, gives:
$$
v(t+1) - v(t) = -\epsilon v(t) + f(t) −{f(t)\epsilon \over 2} + O(\epsilon^2)
$$
then, to get to the desired equation we would have to neglect $\epsilon \over 2$ but leave $\epsilon$. I'm not entirely sure this is correct, or why it should be like this. But it would then yield:
$$
v(t+1) - v(t) = -\epsilon v(t) + f(t) + O\left({\epsilon \over 2} \right)
$$
The term $ v(t+1) - v(t) $ is discrete derivative, i.e. think of the difference quotient from first year calculus with $ h = \Delta t = (t+1)-t = 1 $.
Thus, 
$$
\frac{dv}{dt} = -\epsilon v(t) + f(t)
$$
A: The discretization comes from the solution to the full first-order ODE. You have
$$ v' + \epsilon v = f$$
Using an integrating factor, the general solution to this has the form
$$ v(t) = e^{-\epsilon t} \left( v(0) + \int_0^{t} e^{\epsilon \tau} f(\tau) \, d\tau \right) $$
For small time, we might approximate $f$ as a constant, in which case the above solution becomes
$$ v(t) \approx e^{-\epsilon t} v(0) + \frac{f(0)}{\epsilon} \left(1 - e^{-\epsilon t} \right) $$
Your discretization is a time shift of the above. 
