A tank contains 100 L of pure water. Brine that contains 0.1 kg of salt per liter enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank after 6 minutes?

Let [tex]y(t)[/tex] be the amount of salt (in kg) after [tex]t[/tex] minutes. Then [tex]y(0) = 0[/tex]. The amount of liquid in the tank is 100 L at all times, so the concentration at time [tex]t[/tex] (in minutes) is [tex]y(t) / 1000\ \mathrm{kg/L)[/tex] and

[tex]\dfrac{dy}{dt} = \left( 0.1\dfrac{\mathrm{kg}}{\mathrm{L}}\right) \left( 10 \dfrac{\text{L} }{\text{ min}}\right)- \left[ \dfrac{y(t)}{100} \dfrac{\mathrm{kg}}{\mathrm{L}}\right] \left( 10 \dfrac{\text{L} }{\text{ min}}\right) \\ \\ \dfrac{dy}{dt} = 1 \dfrac{\mathrm{kg}}{\mathrm{min}}-\dfrac{y(t)}{10}\dfrac{\mathrm{kg}}{\mathrm{min}} \\ \\ \dfrac{dy}{dt} = 1 - \dfrac{y}{10} = \dfrac{10 - y}{10} \\ \displaystyle\int \dfrac{dy}{10 - y} = \int \frac{1}{10} dt \\ -\ln|10-y| = \frac{t}{10} + C[/tex]

and [tex]y(0) = 0[/tex] ⇒ [tex]-\ln|10-0| = \frac{0}{10} + C\ \Rightarrow\ C = -\ln 10[/tex]

[tex]-\ln|10-y| = \frac{t}{10} - \ln 10 \\ \Rightarrow \ln|10-y| = \ln 10 - \frac{t}{10} \\ \Rightarrow |10 - y| = e^{\ln 10 - t/10} \\ \Rightarrow |10 - y| = 10e^{- t/10} \\ \Rightarrow 10 - y = \pm 10e^{- t/10} \\ \Rightarrow y = 10 \pm 10e^{- t/10}[/tex]

Choose (-) because that satisfies [tex]y(0) = 0[/tex]

[tex]y(t) = 10 - 10e^{- t/10} \\ y(6) = 10 - 10e^{-6/10} \approx 4.512\ \mathrm{kg}[/tex]

4.512 kg