Beginning at time t=0, fresh water is pumped at the rate of 3 gal/min into a 60-gal tank initially filled with brine. The resulting less-and-less salty mixture overflows at the same rate into a second 60-gal tank that initially contained only pure water, and from there it eventually spills onto the ground. Assuming perfect mixing in both tanks, when will the water in the second tank taste saltiest? And exactly how salty will it then be, compared with the original brine?

It can view the tank as a compartment containing salt. If let x(t) denote the mass of salt in the tank at a time t, it can determine the concentration of salt in the tank by dividing x(t) by the volume of fluid in the tank at a time t. It uses the mathematical model described by the following equation to solve forx(t),

$\frac{dx}{dt}=\mathrm{Inputrate}--\mathrm{Output} \mathrm{rate}..................\left(1\right)$

First, it must determine the rate at which salt enters the first tank. It is given that freshwater flows into the tank at a rate of3 gal/min, which is initially filled with brine. So, It concludes that the input rate of salt into the first tank is, $\left(3gal/min\right)\left(0\right)\mathbf{=}\mathbf{0}$.

It must now determine the output rate of the solution from the first tank.The resulting less-and-less salty mixture overflows at the same rate into a second 60-gal tank.So, we conclude that the output rate of salt from the first tank is,

$\left(3\right)\frac{\mathbf{x}\left(t\right)}{\mathbf{60}}\mathbf{=}\frac{\mathbf{x}\left(t\right)}{\mathbf{20}}$

Substituting the input and output rates from step 2 and step 3 into the equation (1), it will get the following differential equation,

$\frac{dx}{dt}=0-\frac{x\left(t\right)}{20}$

i.e.,$\frac{dx}{dt}+\frac{x\left(t\right)}{20}=0$ …… (2)

Integrating factor, I.F.= $e^{\int \frac{1}{20}dt}=e^{\frac{1}{20}t}$.

Multiplying both sides of equation (2) by $e^{\frac{1}{20}t}$,

$$\begin{gathered} \left(e^{\frac{1}{20}t}\right)\frac{dx}{dt}+\left(e^{\frac{1}{20}t}\right)\frac{x\left(t\right)}{20}=0 \\ \frac{d}{dt}\left(x·e^{\frac{1}{20}t}\right)=0 \end{gathered}$$

Now, integrating both sides,

Therefore $x·e^{\frac{1}{20}t}=C$ …… (3)

where, C is an arbitrary constant.

First, it must determine the rate at which salt enters the second tank. It is given that resulting solution from the first tank flows into the second tank at a rate of3 gal/min, which is initially filled with pure water. So, it concludes that the input rate of salt into the second tank is, $\left(3\right)\left(x\right)\mathbf{=}\mathbf{3}\mathit{x}$.

Where x is the mass of salt in the first tank which enters the second tank.

It must now determine the output rate of the solution from the second tank.The solution from the second tankeventually spills onto the ground. Let the mass of solution in the second tank be y.So, it concludes that the output rate of salt from the second tank is,$\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathit{y}$.

Now as the mass of both the tanks is60 gal.

Therefore,

$$\begin{gathered} \left(60\right)·\frac{dy}{dt}=3x-3y \\ \frac{dy}{dt}=\frac{x}{20}-\frac{y}{20} \end{gathered}$$

$\frac{dy}{dt}+\frac{y}{20}=\frac{x}{20}$ …… (4)

Integrating factor, I.F.=$e^{\int \frac{1}{20}dt}=e^{\frac{1}{20}t}$

Multiplying both sides of equation (4) by $e^{\frac{1}{20}t}$,

$$\begin{gathered} \left(e^{\frac{1}{20}t}\right)\frac{dy}{dt}+\left(e^{\frac{1}{20}t}\right)\frac{y}{20}=\left(e^{\frac{1}{20}t}\right)\frac{x}{20} \\ \frac{d}{dt}\left(y·e^{\frac{1}{20}t}\right)=\left(e^{\frac{1}{20}t}\right)\frac{x}{20} \end{gathered}$$

Now, using equation (3)

$$\begin{gathered} \frac{d}{dt}\left(y·e^{\frac{1}{20}t}\right)=\left(e^{\frac{1}{20}t}\right)\frac{C{e}^{-\frac{1}{20}t}}{20} \\ \frac{d}{dt}\left(y·e^{\frac{1}{20}t}\right)=\frac{C}{20} \end{gathered}$$

Now, integrating both sides,

$y·e^{\frac{1}{20}t}=\frac{Ct}{20}+C_1$

where, C₁ is an arbitrary constant.

$y=\frac{Ct}{20}·e^{-\frac{1}{20}t}+C_1·e^{-\frac{1}{20}t}$ …… (5)

When t=0,y=0So, from equation (5)

$C_1=0$

So, equation (5) becomes

$y=\frac{Ct}{20}·e^{-\frac{1}{20}t}$

Therefore, the mass of salt after the time, t min is role="math" localid="1664262258656" $\mathit{y}\left(t\right)\mathbf{=}\frac{\mathbf{C}\mathbf{t}}{\mathbf{20}}\mathbf{·}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{20}}\mathbf{t}}$.

The water in the second tank will taste saltiest, when y(t) is maximum,

i.e., when$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{0}$.

$$\begin{gathered} \frac{C}{20}\left(e^{-\frac{1}{20}t}-\frac{t}{20}{e}^{-\frac{1}{20}t}\right)=0 \\ {e}^{-\frac{1}{20}t}-\frac{t}{20}{e}^{-\frac{1}{20}t}=0 \\ 1-\frac{t}{20}=0 \\ \frac{t}{20}=1 \\ t=20min \end{gathered}$$

Hence, the water in the second tank will taste saltiest after 20 minutes.

The water in the second tank will be salty after 20 min. Therefore, whent=20min,

$$\begin{gathered} y\left(20\right)=\frac{C\left(20\right)}{20}·e^{-1} \\ y\left(20\right)=\frac{C}{e} \end{gathered}$$

Hence, the water in the second tank will be role="math" localid="1664262536595" $\frac{\mathbf{1}}{\mathbf{e}}$times salty.