$$ \frac{dx}{dt} = \vec f(x(t), x(t-\tau), x(t-\tau_2), ..., x(t-\tau_n)) $$
Example:
First order DE
$$ \frac{dx}{ct} = -x \rightarrow x(t) = x_0e^{-t}$$
First order DDE:
$$\frac{dx}{dt} = -x(t-1)$$
This is a problem. We cannot use an initial value, we need **an initial history function (IHF)**. This is the behaviour of x(t) defined in an interval $[-\tau_0, 0]$, assuming solution time starts at $t=0$

## Method of Steps
Think of DDE as being a mapping between the past interval and the present interval
$$x(t) = \phi_{i-1}(t)$$ on any interval $[t_i-1,t_i]$
$$\int_{\phi_{i-1}(t)}^{x(t)}dx' = - \int_{t_i}^t \phi_{i-1}(t' - 1)dt'$$
Then after some steps
$$\frac{dx}{dt} = -x(t-1) \rightarrow dx' = -x'(t'-1)dt'$$
And then this can be solved at each interval.
- This gets annoying.
- Need to solve at each interval over and over.
## Stability of DDEs
For ODE: equilibrium point is 0 of derivative
For DDE: dx/dt is still 0