How to derive the steady-state solution for simple two-step reaction with differential equations? Consider this simple two step reaction, a variant of a Michaelis-Menten type of problem, where $\ce{A}$ and $\ce{B}$ reversibly bind to make $\ce{AB}$, and $\ce{AB}$ and $\ce{C}$ reversibly bind to make $\ce{ABC}$:

\begin{align}
\ce{A + B  &amp;&lt;=&gt; AB}\\
\ce{AB + C &amp;&lt;=&gt; ABC}
\end{align}

Assume all the rate constants are equal to $k$ for simplicity, and that $[\ce{A}]_0$, $[\ce{B}]_0$, $[\ce{C}]_0$ are initial values for $\ce{A}$, $\ce{B}$, $\ce{C}$, and $[\ce{AB}]_0 = [\ce{ABC}]_0 = 0$. 

What is the proper way to derive the steady-state concentration of $\ce{ABC}$ as a function of initial values? here is an attempt to get $\mathrm{d}[\ce{AB}]/\mathrm{d}t$:
$$\frac{\mathrm{d}[\ce{AB}]}{\mathrm{d}t} = ([\ce{A}][\ce{B}]k + [\ce{ABC}]k) - ([\ce{AB}]k + [\ce{AB}][\ce{C}]k)$$

The first parenthesis of the sum are the two ways to make $\ce{AB}$ and second parenthesis are the two ways to lose $\ce{AB}$. 
We can set $\mathrm{d}[\ce{AB}]/\mathrm{d}t = 0$ and try to solve for $[\ce{ABC}]$, but I am not sure how this helps and where initial values come in.
Based on the discussion in comments, the OP is interested in the equilibrium concentration of the species ABC in this reaction scheme:

$$\ce{A + B + C &lt;=&gt;[k_1][k_2] AB + C &lt;=&gt;[k_3][k_4] ABC}$$

Equilibrium is reached when the rates of the forward and backward reactions of each step are equal. Thus, we have have two equations describing the equilibrium:

$$k_1\ce{[A][B]}=k_2\ce{[AB]}$$
$$k_3\ce{[AB][C]}=k_4\ce{[ABC]}$$

We also know from mass conservation that
$$\ce{[A]_o}=\ce{[A] + [AB] + [ABC]}$$
$$\ce{[B]_o}=\ce{[B] + [AB] + [ABC]}$$
$$\ce{[C]_o}=\ce{[C] + [ABC]}$$

These five equations can be manipulated to find an expression for $\ce{[ABC]}$ in terms of $\ce{[A]_o}$, $\ce{[B]_o}$, and $\ce{[C]_o}$.