# What is the difference between initial conditions and input of system?

Find $$~y_0 (t)~$$, the zero-input component of the response, for a LTI system described by the following differential equation: $$(D^2+6D+9)y(t)=(3D+5)x(t)$$ when the initial conditions are, $$y_0 (0)=3, ~~~~~~~~y ̇_0 (t)=-7~.$$

The zero-input response is: $$y_0 (t)=3e^{-3t}+2te^{-3t}$$

What do the initial conditions physically represent?

If we are given initial conditions, how is this different to stating that there is an initial input at $$~t=0^+~~?~$$

Isn’t the existence of initial conditions implying that there is a forced-input acting on the system?

With initial conditions it will cause transient adjustment to the internal system (homogeneous equation), but in reality how does this occur if we are treating it as having no inputs or outputs during this time?

How can I conceptually distinguish input to the system from the initial conditions of the system?

## 1 Answer

The input to a system can be a time-varying signal, whereas the initial condition is the situation before the input is applied.

Think of a child's swing. The initial condition could be "hanging straight down" ($$\theta = 0$$), while the input could be a force such as $$F(t) = \sin 5 t$$, which might go on forever. Moreover, the initial condition could be of a different nature than the input. Here the initial condition is an angle. The force might actually be any particular value at $$t = 0$$. But the input is the applied force over time.

• This significantly helps me, thank you! I hadn't considered that there would be ongoing 'effects' on the system from a previous input not represented by the given equation. The initial conditions allows one to incorporate this! :) – ndyson0 Aug 21 '19 at 1:21