A large tank contains $81\;\rm{gal}$ of brine in which $20\;\rm{lb}$ of salt are dissolved. Brine containing $3\;\rm{lb}$ of dissolved salt per gallon runs into the tank at the rate of $5\;\rm{gal}/\rm{min}$. The mixture, kept uniform by stirring, drains from the tank at the rate of $2 \;\rm{gal}/\rm{min}$. How much salt is in the tank at the end of $37 \;\rm{min}$? See Figure 6.

Figure 6

Solution Let $y(t)$ be the amount of salt in the tank at time $t$ . The rate of change $\dfrac{dy}{dt}$ of salt in the tank at time $t$ is the rate at which salt enters the tank minus the rate at which salt leaves the tank. Salt enters the tank at the rate of $\left( 3\;\rm{lb}/\rm{gal}\right) \left(5\;\rm{gal}/\rm{min}\right) =15\;\rm{lb}/\rm{min}$.

To find the rate at which the salt exits the tank, we first need to find the concentration of salt in pounds per gallon at time $t$. $$\begin{equation*} \hbox{Concentration}=\dfrac{\hbox{amount of salt}}{\hbox{number of gallons}}= \frac{y(t)}{81+(5-2)t}=\dfrac{y( t) }{81+3t}\;\rm{lb}/\rm{gal} \end{equation*}$$

Then the rate at which the salt exits the tank is $$\begin{equation*} \left( \dfrac{y( t) }{81+3t}\;\rm{lb}/\rm{gal}\right) \left( 2 \;\rm{gal}/\rm{min}\right) =\dfrac{2y( t) }{81+3t}\;\rm{lb}/\rm{min} \end{equation*}$$

The mixture problem is modeled by the differential equation $$\begin{eqnarray*} \dfrac{dy}{dt}=\hbox{rate in }-\hbox{rate out}&=&15-\frac{2y( t) }{ 81+3t}\\ \dfrac{dy}{dt}+\frac{2}{3( 27+t) }y&=&15 \end{eqnarray*}$$

This is a first-order linear differential equation. We use the integrating factor: $$\begin{eqnarray*} e^{\int \!P( t)\,dt}&=&\exp \!\left[ \int \!\dfrac{2}{3( 27+t) } dt\right] =\exp \!\left[ \dfrac{2}{3}\ln ( 27+t) \right]\\ &=&\exp\,[ \ln ( 27+t) ^{2/3}] =(27+t)^{2/3} \end{eqnarray*}$$

We multiply the differential equation by the integrating factor to obtain $$\begin{eqnarray*} (27+t)^{2/3}\dfrac{dy}{dt}+\dfrac{2y}{3( 27+t) ^{1/3}} &=&15(27+t)^{2/3} \\ \frac{d}{dt}[y(t)(27+t)^{2/3}] &=&15(27+t)^{2/3} \end{eqnarray*}$$

Now we integrate both sides with respect to $t$.

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Next we use the initial condition that $y( 0) =20\;\rm{lb}$. Then $$\begin{eqnarray*} y( 0) &=&20=9(27)+C(27)^{-2/3}\\ -223 &=&\dfrac{C}{9} \\ C &=&-2007 \end{eqnarray*}$$

The amount of salt $y( t)$ in the brine at time $t$ min is $$\begin{equation*} y(t)=9(27+t)-2007(27+t)^{-2/3} \end{equation*}$$

The amount of salt in the tank at the end of $37\;\rm{min}$ is $$y(37)=9(27+37)-2007(27+37)^{-2/3}\approx 451\;\rm{lb}$$