Assuming that the bimolecular chemical reaction $\ce{A + B ->[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs
$$\begin{array}{rl} \dot a &= - \kappa \, a \, b\ \dot b &= - \kappa \, a \, b\end{array}$$
where $\kappa > 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain
$$a (t) - b (t) = a_0 - b_0$$
where $a_0 > 0$ and $b_0 > 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0 - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows
$$\dot a = - \kappa \, a \, \left( a - (a_0 - b_0) \right)$$
which can be rewritten in the form
$$\frac{\mathrm d a}{a \, \left( a - (a_0 - b_0) \right)} = - \kappa \, \mathrm d t$$
Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion
$$\left( \frac{1}{a - (a_0 - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0 - b_0) \, \mathrm d t$$
Integrating, we obtain
$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a_0 - (a_0 - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0 - b_0) \, t$$
which can be rewritten as follows
$$\ln \left( \frac{a (t) - (a_0 - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0 - b_0) \, t$$
Exponentiating both sides, we obtain
$$\frac{a (t) - (a_0 - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)$$
and, eventually, we obtain
$$\boxed{\begin{array}{rl} &\ a (t) &= \dfrac{a_0 - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\\ b (t) &= \dfrac{(a_0 - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0 - b_0) \, t)}\ & \end{array}}$$
Taking the limit,
$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 & \text{if } a_0 > b_0\\ 0 & \text{if } a_0 < b_0\end{cases}$$
$$\$$
$$\lim_{t \to \infty} b (t) = \begin{cases} 0 & \text{if } a_0 > b_0\\ b_0 - a_0 & \text{if } a_0 < b_0\end{cases}$$
What if $a_0 = b_0$?
Previously, we assumed that $a_0 \neq b_0$. 374