Problem source

A liquid of fixed volume $V$ is made up of two chemicals $A$ and $B\,$. A reaction takes place in which $A$ converts to $B\,$. The volume of $A$ at time $t$ is $xV$ and the volume of $B$ at time $t$ is $yV$ where $x$ and $y$ depend on $t$ and $x+y=1\,$. 
The rate at which $A$ converts into $B$ is given by $kVxy\,$, where $k$ is a  positive constant. Show that if both $x$ and $y$ are strictly positive at the start, then at time $t$
\[
y= \frac {D\e^{kt}}{1+D \e^{kt}} \;,
\]
where $D$ is a constant. 
Does $A$ ever completely convert to $B\,$? Justify your answer.

Solution source

We have $\dot{A} = -kVxy$ or $\dot{x}V = -kVxy$, ie $\dot{x} = -kxy$ and similarly $\dot{y} = kxy = k(1-y)y$.

\begin{align*}
&& \frac{\d y}{\d t} &= ky(1-y) \\
\Rightarrow && \int k \d t &= \int \frac{1}{y(1-y)} \d y \\
\Rightarrow && kt &= \int \left ( \frac{1}{y} + \frac{1}{1-y} \right) \d y \\
&&&= \ln y - \ln (1-y) + C\\
\Rightarrow && kt &= \ln \frac{y}{D(1-y)} \\
\Rightarrow && De^{kt} &=  \frac{y}{1-y} \\
\Rightarrow && y(1+De^{kt}) &= De^{kt} \\
\Rightarrow && y &= \frac{De^{kt}}{1+De^{kt}}
\end{align*}

As $t \to \infty$ $y \to \frac{D}{D} = 1$ so depending on how fine grained we want to go we might say that 'yes it completely converts' when there is an immeasurably small amount of $A$ left, or we might say it doesn't since it only tends to $1$ and never actually reaches it.