Critically damped system Coming from engineering background, my math might not be as rigour here. Please enlighten me.

By definition, critically damped 2nd order system ($\zeta=1$) does not have overshoot.

Is the statement above (my assumption) false?
For context, this is the system:
$$
Q\ddot{x}+2\zeta \omega_n\dot{x}+\omega_n^2x=\omega_n^2r
$$
which is also in the form:
$$
Q\ddot{x}+a\dot{x}+bx=br
$$
where $r$ is some constant.
I know that the response is in the form:
$$
x(t)=r\left[1−(k_1+k_2t)e^{\frac{−\omega_n}{Q}t}\right]
$$
Where $k_1$, $k_2$ is some constant depend on initial condition.
For $x(0) = 0$, We know that $k_1=0$, but how to determine $k_2$ (which depends on $x'(0)$)?
I know that this might be very trivial for some of you, but I'm really confused.
For context, I asked a similar question in the electrical engineering site. If you are interested, the problem is the 2nd order system that I set "gains" for (gains correspond to changing the coefficient in the ode (a & b)), behave critically damped. However, when I set up a the original first order system with the input of the controlled gains, it had an overshoot. If you know why kindly please enlighten me as well.
This is the unstable first order system trying to be controlled:
$$
p\dot{x}-lx=r
$$
Letting
$$
r= K_p\left(c-x\right) + K_i\int _0^tc-x\ dt\ + K_d\left(\dot{c} - \dot{x}\right)
$$
This is a PID control for 1st order system. Making the sytem to become 2nd order. and track the value $c$
 A: OK, there's a lot to unpack here, so let's take this one step at a time.
ODE solution
The ODE
$$
Q\ddot{x}+2\zeta\omega_n\dot{x}+\omega_n^2x=\omega_n^2r
$$
has the characteristic equation
$$
Q\lambda^2 + 2\zeta\omega_n\lambda + \omega_n^2 = 0
$$
and eigenvalues
$$
\lambda \in \left\{{\omega_n \over Q}\left(-\zeta + \sqrt{\zeta^2 - Q}\right), {\omega_n \over Q}\left(-\zeta -\sqrt{\zeta^2 - Q} \right)  \right\}
$$
so it can only be critically damped if $\zeta^2 = Q$. For  the sake of argument, let's assume it is so that
$$
\lambda = -{\omega_n\zeta \over Q} = -{\omega_n \over \zeta}
$$
and $x$ has the homogeneous solution
$$
x_h = (k_1 + k_2t)e^{-t\omega_n/\zeta}.
$$
For the particular solution, since the input $\omega_n^2 r$ is constant, so is the steady-state response, therefore $x_p=C$. The complete solution is
$$
x = x_p + x_h = C + (k_1 + k_2t)e^{-t\omega_n/\zeta}
$$
with first derivative
$$
\dot{x} = \left[k_2 - {\omega_n \over \zeta}(k_1 + k_2t) \right]e^{-t\omega_n/\zeta}
$$
In control problems, it's usual to assume that the system has no energy stored at $t=0$ unless you have information indicating otherwise, so $x(0)=\dot{x}(0)=0$. From the initial conditions, we have $k_1 = -C$ and
$$
k_2 = {\omega_n \over \zeta}k_1 = -{\omega_n \over \zeta}C.
$$
To find $C$, we notice that $x = C$ and $\dot{x}=\ddot{x}=0$ for $t\to\infty$, so, if we were to plug our candidate solution and its derivatives back into the ODE and evaluate it for $t\to\infty$ we'd get $\omega_n^2C = \omega_n^2r$, hence
$$
C = r\quad k_1 = -r\quad k_2 = -{\omega_n \over \zeta}r
$$
which gives the final solution
$$
x = r \left[ 1 - \left(1 + {\omega_n \over \zeta}t\right)e^{-t\omega_n/\zeta}\right].
$$
Having said all that, I believe you may have gotten your coefficients mixed up. If you want $\zeta$ to represent the damping ratio, your ODE has to be expressed in the form $\ddot{x} + 2\zeta\omega_n\dot{x} + \omega_n^2x = f(x)$.
System control
Now, for the control problem. You said you're trying to control a 1st-order unstable plant with a PID controller. Your controlled system then has a closed-loop transfer function of the form (assuming unity feedback)
$$
H(s) = {(s^2K_d + sK_p + K_i)/p \over s^2(1 + K_d/p) + s[(K_p-l)/p] + K_i/p},
$$
where
$$
\omega_n^2 = {K_i/p \over 1 +K_d/p} = {K_i \over p + K_d}
$$
and
$$
2\zeta\omega_n = {(K_p - l)/p \over 1 + K_d/p} = {K_p - l \over p + K_d}.
$$
If all you care about is having a critically damped system, you can find, after a few manipulations, the constraint
$$
2K_i^2 = (p+K_d)(K_p-l).
$$
If you also need a specific natural frequency $\omega_n$, it can provide you an extra constraint that helps narrow down the possible values for $K_p$, $K_i$ and $K_d$. Of course, you have to make sure the closed-loop poles are stable.
