Suppose that the initial state $\omega_0$ is normally distributed with mean $\mu_0$ and variance $\sigma_0^2$. The state then evolves according to the stochastic differential equation
$$\tag{*}d\omega_t=\kappa\omega_tdt+\sigma dZ_t$$
where the driving process $\{Z_t\}_{t≥0}$ is a standard Brownian motion, independent of the initial state $\omega_0$. The variance rate $\sigma^2$ is strictly positive. The percentage drift $\kappa$ has unrestricted sign. If $\kappa < 0$, then the state follows a mean-reverting Ornstein–Uhlenbeck process. If $\kappa = 0$, then $\omega_t − \omega_0 = \sigma Z_t$, so the state follows a Brownian motion with zero drift. If $\kappa > 0$, then the state process is explosive. The solution of $(*)$ is given by $$\omega_t=\omega_0 e^{-\kappa t} +\sigma\int_{0}^te^{\kappa(s-t)}dZ_s$$
The effect of $\kappa$ can be seen in the formula for the conditional expectation $\mathbb{E}[\omega_s|\omega_t]$.
Can we show that for any $s > t ≥ 0$, we have $\mathbb{E}[\omega_s|\omega_t] =e^{\kappa(s−t)}\omega_t$ and why?
Take the expected value of both sides of your equation, since the expected value of the Brownian motion is zero, you have $$ \frac{d\langle \omega_t\rangle}{dt} = \kappa \langle\omega_t\rangle. $$ which you can integrate from $t$ to $s$ given the initial condition at $\omega_t$ to obtain your result.