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Use differential equations to determine the salinity of a solution

by RoRi

This exercise refers to the previous exercise (Section 8.7, Exercise #9). Given a tank with salt in the bottom dissolving into the liquid of the tank at a rate proportional to the difference between the concentration of salt in the solution and the concentration of salt in a completely saturated solution (3 pounds of salt per gallon). If the water in the tank were pure, 1 pound of salt would dissolve per minute. How much salt will be in the solution at the end of one hour?

Let y = f(t) = number of pounds of salt at time t. Then y' is the salt added by the dissolving substance minus the salt removed by run off. The salt added by the dissolving substance is

    \[ k \left( 3 - \frac{f(t)}{100 + t} \right), \qquad 3k = 1 \implies k = \frac{1}{3}. \]

Combining this with the equations we derived in the previous exercise (linked above) we have

    \[ y' = -2 \left( \frac{f(t)}{100+t} \right) + 1 - \left( \frac{f(t)}{3(100+t)} \right) \]

    \[ \implies y' + \frac{7}{3(100+t)} y = 1 \qquad \text{with} \quad f(0) = 50. \]

Therefore, using Theorem 8.3 (page 310 of Apostol) to solve this first-order linear differential equation we have

    \begin{align*} f(t) &= 50 \cdot \frac{100^{\frac{7}{3}}}{(100+t)^{\frac{7}{3}}} + \frac{1}{(100+t)^{\frac{7}{3}}} \int_0^t (100+x)^{\frac{7}{3}} \, dx \\[9pt] &= 50 \cdot \frac{100^{\frac{7}{3}}}{(100+t)^{\frac{7}{3}}} + \frac{1}{(100+t)^{\frac{7}{3}}} \left( \frac{3}{10} (100+t)^{\frac{10}{3}} - \frac{3}{10} (100)^{\frac{10}{3}} \right) \\[9pt] &= 50 \cdot \frac{100^{\frac{7}{3}}}{(100+t)^{\frac{7}{3}}} + \frac{3}{10} (100+t) - \frac{3}{10} \cdot \frac{100^{\frac{10}{3}}}{(100+t)^{\frac{7}{3}}}. \end{align*}

So for t = 60 we have

    \[ f(60) = 54.7 \text{ pounds of salt}. \]

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