Why is the right side called the "input?"

[maxbashi]

I'm looking at linear, 1st order ODEs, like

y' + p(t)x = q(t)

The notes I'm looking at are calling q(t) the "input" of the system, but I'm not sure why. I understand how to solve the equations, but I must be looking at it differently or something. To quote the notes:

The left hand side represents the SYSTEM.
The right hand side represents an outside influence on the system: it's a "signal," the "input signal." A "signal" is just a function of time.
The system responds to the input signal and yields the function x(t), "output signal."

I guess I'm just not intuitively understanding the way the author is looking at ODEs. Any help?

[Mark44]

You have too many variables in your differential equation. It almost certainly should be
y' + p(t) y = q(t).

The related homogeneous equation is y' + p(t)y = 0. By separating variables, the solution can be found to be
|y| = Ae^{-\int p(t) dt}

If you are given the initial condition, y(t0) = y0, you can determine A. The function p(t) will determine the behavior of this solution. If this differential equation is modeling a physical process, the typical long-term behavior of this solution is that y(t) --> 0 as t --> infinity.

For the nonhomogeneous equation, y' + p(t)y = q(t), the function on the right side produces a different solution with two parts: the transient solution (which dies out in time), and the steady-state solution. In a sense, q(t) is "driving" the system and can be thought of as acting as an input to the system.

[jambaugh]

Let me add that the terminology comes not from the theory of the ODE itself but from some specific typical application. Presumably the ODE is from a model for the dynamics of some physical system. Then the q(t) probably represents an external (time dependent) force.

For example, a mass-spring system has force proportional to position:
F = -kx
So from Newton's laws F=ma:
m \ddot{x}= -kx + F_{ext}
or
m\ddot{x}+ kx = F_{ext}(t)