{#subsec:SS}
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Equations are suitable for calculating the steady states of this system. For a steady state $\lim_{t\rightarrow\infty} P(1,t | k_i)=P^*(1 | k_i)$ applies. Using and integration by parts gives the following coupled implicit equations for the steady states $$\begin{aligned}
    {P^*(1|k_1)}&= \frac{\tau{\gamma_{k_1}}^*}{1+\tau{\gamma_{k_1}}^*},\qquad{P^*(1|k_2)}= \frac{\tau{\gamma_{k_2}}^*}{1+\tau{\gamma_{k_2}}^*}.
    \label{eq:SS_rbnet}\end{aligned}$$ The values of ${P^*(1|k_1)}^*$ and ${P^*(1|k_2)}^*$ define the stationary order in the two subpopulations. In Eq. we introduced the mean relaxation time of the excited state $\tau=\int_0^\infty
t\,w_1(t)\mathrm{d} t$ and the steady state activation rate $\gamma^*_{k_i}=\gamma[k_i r^*]$ for $i=1,2$ depending on the steady state mean-field $r^*$. Equation defines the order parameter of the full network. It becomes $r^*=\left\langle k P^*(1 |
  k)\right\rangle / \langle k \rangle $. Taken at steady state, this yields a transcendent equation for the steady state value $r^*$ of the mean-field which yields $$r^*= \frac{1}{\langle k \rangle}\left\langle \frac{k}{1+\frac{1}{\tau\gamma_{k}^*}}\right\rangle.
\label{eq:rSS_rbnet}$$ This equation can possess several solutions which we will discuss in detail, later on. In case of large noise $D \to \infty$, only the homogeneous disordered solution $r^*=1/2$ exists. In this limit the exponential function becomes unity and since we will select $\gamma_0
\tau \approx 1$ the disordered state is characterized by $r^*=1/2$. A constant value of $r^*$ does not imply that the 