A tank originally contains 100 gallon of fresh water. Then water containing 0.5 Lb of salt per gallon is pourd into the tank at a rate of 2 gal/minute, and the mixture is allowed to leave at the same rate. After 10 minute the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at end of an additional 10 minutes.

Let [tex]S(t)[/tex] denote the amount of salt (in lbs) in the tank at time [tex]t[/tex] min up to the 10th minute. The tank starts with 100 gal of fresh water, so [tex]S(0)=0[/tex].

Salt flows into the tank at a rate of

[tex]\left(0.5\dfrac{\rm lb}{\rm gal}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = 1\dfrac{\rm lb}{\rm min}[/tex]

and flows out with rate

[tex]\left(\dfrac{S(t)\,\rm lb}{100\,\mathrm{gal} + \left(2\frac{\rm gal}{\rm min} - 2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

Then the net rate of change in the salt content of the mixture is governed by the linear differential equation

[tex]\dfrac{dS}{dt} = 1 - \dfrac S{50}[/tex]

Solving with an integrating factor, we have

[tex]\dfrac{dS}{dt} + \dfrac S{50} = 1[/tex]

[tex]\dfrac{dS}{dt} e^{t/50}+ \dfrac1{50}Se^{t/50} = e^{t/50}[/tex]

[tex]\dfrac{d}{dt} \left(S e^{t/50}\right) = e^{t/50}[/tex]

By the fundamental theorem of calculus, integrating both sides yields

[tex]\displaystyle S e^{t/50} = Se^{t/50}\bigg|_{t=0} + \int_0^t e^{u/50}\, du[/tex]

[tex]S e^{t/50} = S(0) + 50(e^{t/50} - 1)[/tex]

[tex]S = 50 - 50e^{-t/50}[/tex]

After 10 min, the tank contains

[tex]S(10) = 50 - 50e^{-10/50} = 50 \dfrac{e^{1/5}-1}{e^{1/5}} \approx 9.063 \,\rm lb[/tex]

of salt.

Now let [tex]\hat S(t)[/tex] denote the amount of salt in the tank at time [tex]t[/tex] min after the first 10 minutes have elapsed, with initial value [tex]\hat S(0)=S(10)[/tex].

Fresh water is poured into the tank, so there is no salt inflow. The salt that remains in the tank flows out at a rate of

[tex]\left(\dfrac{\hat S(t)\,\rm lb}{100\,\mathrm{gal}+\left(2\frac{\rm gal}{\rm min}-2\frac{\rm gal}{\rm min}\right)t}\right) \left(2\dfrac{\rm gal}{\rm min}\right) = \dfrac{\hat S(t)}{50} \dfrac{\rm lb}{\rm min}[/tex]

so that [tex]\hat S[/tex] is given by the differential equation

[tex]\dfrac{d\hat S}{dt} = -\dfrac{\hat S}{50}[/tex]

We solve this equation in exactly the same way.

[tex]\dfrac{d\hat S}{dt} + \dfrac{\hat S}{50} = 0[/tex]

[tex]\dfrac{d\hat S}{dt} e^{t/50} + \dfrac1{50}\hat S e^{t/50} = 0[/tex]

[tex]\dfrac{d}{dt} \left(\hat S e^{t/50}\right) = 0[/tex]

[tex]\hat S e^{t/50} = \hat S(0)[/tex]

[tex]\hat S = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-t/50}[/tex]

After another 10 min, the tank has

[tex]\hat S(10) = 50 \dfrac{e^{1/5}-1}{e^{1/5}} e^{-1/5} = 50 \dfrac{e^{1/5}-1}{e^{2/5}} \approx \boxed{7.421}[/tex]

lb of salt.

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