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\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}
{\bf Question}
Two tanks both initially contain 200 L of fresh water. Starting at
$t=0$ brine containing 5 kg/L of salt is added to the first tank
at the rate of 2 L/min. This first tank is continually stirred.
The uniform solution from the first tank is transferred to the
second tank at the rate of 2 L/min. This second tank is also
stirred. A uniform mixture leaves the second tank also at a rate
of 2 L/min. What is the concentration of the mixture leaving the
first tank at time $t$? What is the concentration of the mixture
leaving the second tank at time $t$?
\vspace{0.25in}
{\bf Answer}
$\ds x(t)=$ salt in tank 1 $\ds \hspace{0.5in} x(0)=0$
$\ds y(t)=$ salt in tank 2 $\ds \hspace{0.5in} y(0)=0$
The water balance is such that both tanks contain 200 litres at
all times.
Salt balance in tank 1:
$\ds \begin{array}{cccccc} rate\ of\ change &=& rate\ of\ salt\ in
&-& rate\ of\ salt\ out\\ of\ salt\\ \frac{dx}{dt} &=&
5*2&-&\frac{x}{200}*2\\ \frac{dx}{dt} &=& 10 &-& \frac{x}{100} &
(1)\end{array}$
Salt balance in tank 2:
$\ds \begin{array}{cccccc} rate\ of\ change &=& rate\ of\ salt\ in
&-& rate\ of\ salt\ out\\ of\ salt\\ \frac{dy}{dt} &=&
\frac{x}{200}*2 &-&\frac{y}{200}*2\\ \frac{dy}{dt} &=&
\frac{x}{100} &-& \frac{y}{100} & (2)\end{array}$
Solve (1) as a linear equation with $\ds x(0)=0$, and we get
$$ x(t)=1000\left(1-e^{-\frac{t}{100}}\right)$$
Put this solution into (2) and solve as a linear equation with
$\ds y(0)=0$, and we get
$$
y(t)=1000\left(1-e^{-\frac{t}{100}}-\frac{te^{-\frac{t}{100}}}{1000}\right)$$
So the concentration is $\ds \frac{x}{100}$ and $\ds
\frac{y}{100}$.
\end{document}