Problem source

In the manufacture of Grandma's Home Made Ice-cream, chemicals $A$ and $B$ pour at constant rates $a$ and $b-a$ litres per second ($0 < a < b$) into a mixing vat which mixes the chemicals rapidly and empties at a rate $b$ litres per second into a second mixing vat.
At time $t=0$ the first vat contains $K$ litres of chemical $B$ only. Show that the volume $V(t)$ (in litres) of the chemical $A$ in the first vat is governed by the differential equation 
\[
\dot{V}(t)=-\frac{bV(t)}{K}+a,
\]
and that 
\[
V(t)=\frac{aK}{b}(1-\mathrm{e}^{-bt/K})
\]
for $t\geqslant0.$
The second vat also mixes chemicals rapidly and empties at the rate of $b$ litres per second. If at time $t=0$ it contains $L$ litres of chemical $C$ only (where $L\neq K$), how many litres of chemical $A$ will it contain at a later time $t$?

Solution source

The total volume in the first vat at time $t$ is always $K$, since $b$ litres per second are coming in and $b$ litres per second are going out.

\begin{align*}
&&\frac{\d V}{\d t} &= \underbrace{a}_{\text{incoming chemical }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{\frac{V(t)}{K}}_{\text{fraction of outgoing which is }A} \\
&&&= a - b \frac{V}{K} \\
\Rightarrow && \int \frac{1}{a-b\frac{V}{K}}\d V &= \int \d t \\
&& - \frac{K}{b} \ln |a - b \frac{V}{K}| &= t +C\\
(t,V) = (0,0): && -\frac{K}{b} \ln a &= C \\
\Rightarrow && 1-\frac{b}{a} \frac{V}{K} &= e^{-bt/K} \\
\Rightarrow && V &= \frac{aK}{b} (1 - e^{-bt/K})
\end{align*}

\begin{align*}
&&\frac{\d W}{\d t} &= \underbrace{b}_{\text{incoming volume}} \cdot \underbrace{\frac{a}{b}  (1 - e^{-bt/K})}_{\text{incoming fraction }A} - \underbrace{b}_{\text{outgoing volume}} \cdot \underbrace{ \frac{W(t)}{L}}_{\text{fraction of outgoing which is }A} \\
&&&= a (1 - e^{-bt/K}) - b \frac{W}{L} \\
\Rightarrow && \frac{\d W}{\d t} + \frac{b}{L} W &= a (1-e^{-bt/K}) \\
&& \frac{\d}{\d t} \left ( e^{b/L t} W\right) &= ae^{b/L t}(1-e^{-bt/K}) \\
\Rightarrow && W &= e^{-bt/L} \left ( \frac{aL}{b}e^{b/Lt} - \frac{a}{\frac{b}{L} - \frac{b}{K}}e^{b/L t - b/K} \right) + Ce^{-bt/L} \\
&&&= \frac{aL}{b} \left (1 - \frac{K}{K-L}e^{-b/Kt} \right)+ Ce^{-bt/L} \\
(t,W) = (0,0): && 0 &= \frac{aL}{b} \frac{-L}{K-L} + C \\
\Rightarrow && C &= \frac{aL^2}{b(K-L)} \\
\Rightarrow && W &= \frac{aL}{b} \left (1 - \frac{K}{K-L} e^{-bt/K} + \frac{L}{K-L} e^{-bt/L}  \right)
\end{align*}