Steady-state solution and initial conditions Let's say that we have the following first order differential equation:
$$\frac{d \rho(t)}{d t}=F(\rho(t))$$
with some given initial condition $\rho(0)$.
I am interested in the steady-state solution $\rho_{ss}$, and to find it we set $\left.\frac{d\rho(t)}{d t}\right|_{t_{ss}}=0$ and solve the algebraic equation 
$$F(\rho_{ss})=0$$
However, how in this situation do we account for the initial condition $\rho(0)=\rho_0$, without solving the full differential equation? After solving the algebraic equation I get infinetely many solutions (actually two distinct solutions, but their superposition is also a solution).
P.s. I don't know if this helps, but in my case there is a conserved quantity $Q$, which is the same for $\rho(0)$ and $\rho_{ss}$.
 A: Solving the equation $\quad F(\rho)=0\quad $ gives $\rho_{ss}$, but doesn't gives the function $\rho(t)$. So we have to find the function $\rho(t)$ in which $\rho_0$ is a parameter.
$$\frac{d \rho(t)}{d t}=F(\rho(t))\quad\to\quad \frac{d\rho}{F(\rho)}=dt$$
$$\int \frac{d\rho}{F(\rho)}= t+c$$
$F(x)$ is a known function. 
Suppose that it is possible to find an antiderivative for $\frac{1}{F(x)}$ , namely $G(x)$ , which then is a known function :
$$G(\rho)=t+c$$
$$\rho=G^{-1}(t+c)$$
where $\quad G^{-1}\quad$ is the inverse function of $G$.
\begin{cases}
\text{Initial point :} \quad G(\rho(0)) =0+c=c \quad\to\quad \rho= G^{-1}\left(t+G(\rho(0)) \right)\\
\text{Steady state point :} \quad \rho_{ss}=G^{-1}(t_{ss}+c)
\end{cases}
This leads to the relationship between $\rho_{ss}$ and $\rho_0=\rho(0)$ :
$$\rho_{ss}=G^{-1}\left(t_{ss}+G(\rho_0)\right)$$
where $G$ and $G^{-1}$ are known functions derived from the given function $F$. 
$t_{ss}$ is a root of the equation $\quad F\left(\rho(t)\right)= F\left( G^{-1}\left(t+G(\rho_0)\right) \right)=0\quad$ where all functions are known.
A: A steady state solution is a constant, that is, time-independent solution to the differential equation. So, if there is a steady state solution (let's call it $\rho_{ss}$), then it must be constant, and it must solve the differential equation. Because $\rho_{ss}$ is constant in time, we have $\frac{\text{d} \rho_{ss}}{\text{d} t} = 0$. Thus, because $\rho_{ss}$ must solve the differential equation, we know that 
\begin{equation}
\frac{\text{d} \rho_{ss}}{\text{d} t} = 0 = F(\rho_{ss}) \tag{1}
\end{equation}
must be satisfied. 
Looking again at equation $(1)$ from a slightly different perspective, we see that once we've found a value $x_*$ for which $F(x_*) = 0$, then $x_*$ is a steady state solution to our differential equation. 
Steady state solutions are independent of time, so they have the same value for all time. So, and this is important, if you take your differential equation and you choose your initial value $\rho(0)$ to be equal to the stationary value $\rho_{ss}$, then the solution will stay constant. 
On the other hand, if you choose an initial value $\rho(0)$ for which $F$ does not vanish, i.e. $F(\rho(0)) \neq 0$, then you will immediately start to move away from the initial value $\rho(0)$, because the initial velocity is nonzero: $\frac{\text{d} \rho}{\text{d} t}(0) = F(\rho(0)) \neq 0$.
So, if you choose your initial condition $\rho(0)$ just right (such that $F(\rho(0)) = 0$), then you've hit a steady state solution, and you'll stay on that steady state solution for all time. 
A: I am having some difficulty understanding the question, but I wanted to add some comments which were not addressed by previous responses.  As I understand you have a 1D ODE: $\rho' =  F(\rho)$ and you want to understand the steady states of $\rho$ as well as how initial conditions affect the steady states.  As you have said, the steady states are the solutions of $F(\rho)=0$, say $\{\rho_i\}_{i\in I}$. I will assume that these $\rho_i$ are somewhat well behaved (countable, isolated, etc).
Then, given any $\rho(0)$, we see that either $\rho(t) \rightarrow \rho_i$ for some $i$ or $\rho(t)\to \pm \infty$. To completely classify the behavior there are two cases:
Case 1: $\rho(0)=\rho_i$. This case is trivial and it is clear that $\rho(t)=\rho_i$ for all $t$.
Case 2: $\rho(0) \neq \rho_i$ for any $i$.  Here it can either happen that $\rho(t)\to \rho_i$, where $\rho_i$ is a stable equilibrium, or $\rho(t)\to \pm \infty$. 
As an example consider $\rho' = \rho^2-1$. Plotting the graph of $\rho^2-1$ on the $\rho'$ vs $\rho$ plane reveals the entire picture (I apologize, I do not know how to include graphics here. I recommend reading Section 2.1 of Strogatz Nonlinear Dynamics and Chaos if you are unfamiliar).  If $\rho(0)<1$ then $\rho(t) \to -1$. If $\rho(0)>1$ then $\rho(t) \to +\infty$. If $\rho(0)=1$ then $\rho(t)=1$ for all time. This is due to stability of the equilibrium point $\rho = -1$ and instability of $\rho=1$. That is, the interaction of initial conditions with equilibrium solutions depends on stability of the equilibria.
