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Given an ode $x' = f(t)$. Then a basic Euler discretization scheme yields $$ x_{n+1} = x_n + h f(t_n).$$

Now suppose you have a delay differential equation, say $x' = f(t-\tau)$, does it make sense to discretize as follows: $$ x_{n+1} = x_n+ hf(t_n - \tau)?$$

Gorg
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    Is this really the question you want to ask? The solution of your ODE is just the integral of $f$. For the delay case not much changes, it is just a shift of the function $f$ and so really still an ODE. The answer by @Canardini is perfect then. But maybe you are more interested in $x' = f(t,x)$ or a delayed version thereof? This would be more interesting. – Fabian Wirth Dec 06 '16 at 11:26
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    @FabianWirth Yes, I am also interested in a delayed versionof $x' = f(t,x)$. Any suggestions? – Gorg Dec 20 '16 at 13:20
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    If you want to ask a different question, then please pose a new one. – Fabian Wirth Dec 20 '16 at 17:04

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We fix $\tau$, and we define the function g, such as $g(t)=f(t-\tau)$ we have to solve $x'(t)=g(t)$.

Discretize it with an Euler scheme ( with all the convergence issues that come with it), it is not different from your initial equation.

Canardini
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    Thanks for your answer. Do you have any suggestions for the delayed version of $x' = f(t,x)$? – Gorg Dec 20 '16 at 13:30