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Use differential equations to find the salinity of a solution

We begin with a tank containing 100 gallons of water and 50 pounds of dissolved salt. Pure water is added to the tank at a rate of 3 gallons per minute (the concentration of the salt is kept constant by stirring). If the mixture runs out of the tank at 2 gallons per minute, how much salt (in pounds) is in the tank at the end of 60 minutes?


Let y = f(t) be the number of pounds of salt in the tank at time t. Since water leaving the tank at a rate of 2 gallons per minute, and at time t there are 100 + (3-2)t = 100 + t gallons of water in the tank (since water is entering the tank at a rate of 3 gallons per minute) we have

    \[ y' = -2 \left( \frac{y}{100+t} \right). \]

(Since y' is the change in the amount of salt at time t and \frac{y}{100+t} is the concentration of salt in the water at time t.) Therefore, we have the first-order differential equation

    \[ y' + \left( \frac{2}{100+t} \right)y = 0. \]

Applying Theorem 8.3 (page 310 of Apostol) with

    \[ P(t) = \frac{2}{100+t}, \qquad Q(t) = 0, \qquad A(t) = \int_0^t \frac{2}{100+x} \, dx = 2 \log(100+t) - 2\log(100), \]

we have

    \begin{align*}  f(t) &= 50 e^{2 \log 100 - 2 \log (100+t)} \\  &= \frac{50 \cdot 100^2}{(100+t)^2}. \end{align*}

Therefore, at time t = 60 we have

    \[ f(60) = \frac{50 \cdot 100^2}{160^2} = 19.5 \text{ pounds of salt}. \]

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