How to derive the steady-state solution for simple two-step reaction with differential equations?
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 $\ce{[C]_o}$.