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.
