Control system with equation C = A*x + B*dx/dt

This question came up in a computer science / robotics exam and I still don't know the solution for it. I figured out that it's classical mechanics related, so I thought this might be the best place to ask it.

Suppose a control system is described by the equation

C = A*x + B*dx/dt


where B is proportional to the mass of the robot. The behaviour of the system can be characterised by the steady state (e.g. the asymptotic velocity of the robot) and the half-life time of the decrease of the distance to the steady state. Explain how the steady state and the half-life change if the mass of the robot is doubled.

I've figured out that the steady state doesn't change, as when dx/dt is 0 B is not affecting the solution.

And this is how far I understand it. Can you explain to me, what kind of movement is this, what is the real-life meaning of the steady state and half-life for this movement and that how to calculate the change in the half-life?

-

This is actually a question for mathematical forum - solving linear non-homogenous differential equations. You have to find solutions for homogenous ($C=0$) and non-homogenous ($C\ne0$) equation. Non-homogenous solution is obviously

$$x_\text{N} = \frac{C}{A},$$

while homogenous solution in

$$x_\text{H} = x_0 \exp[-\frac{A}{B}t].$$

The whole solution of the differential equation is the sum of both solutions $x = x_\text{N} + x_\text{H}$. The point is that for $t \rightarrow \infty$ $x_\text{H} \rightarrow 0$, therefore robot is exponentially getting closer to non-homogenous solution $x_\text{N}$.

I would call $x_\text{N} = \frac{C}{A}$ steady state, while half time $\tau$ can be obtained by equation

$$x_0 \exp[-\frac{A}{B}\tau] = \frac{x_0}{2}.$$

-