Question 11.4.5: (A Mixing Problem) Consider a tank holding 100 gallons of wa...

(A Mixing Problem) Consider a tank holding 100 gallons of water in which is dissolved SO pounds of salt. Suppose that 2 gallons of brine, each containing 1 pound of dissolved salt, run into the tank per minute, and the mixture, kept uniform by highspeed stirring, runs out of the tank at the rate of 2 gallons per minute. Find the amount of salt in the tank at any time t.

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Let x(t) be the number of pounds of salt at the end of t minutes. Since each gallon of brine that enters the tank contains 1 pound of salt, we know that 2 pounds of salt are entering the tank each minute. However, as 2 gallons of solution out of 100 gallons are leaving the tank, (2 / 100) x=0.02 x pounds of salt leave the tank each minute. These data lead to the differential equation

\frac{d x}{d t}=2-0.02 x

or

\frac{d x}{d t}+0.02 x=2

Multiplying both sides of this equation by the integrating factor e^{0.02 t} yields

\frac{d}{d t}\left(x e^{0.02 t}\right)=2 e^{0.02 t}.

We then integrate:

x e^{0.02 t}=\frac{2}{0.02} e^{0.02 t}+C=100 e^{0.02 t}+C

and

x(t)=100+C e^{-0.02 t}.

But we are told that x(0)=50. Then

x(0)=100+C=50,

so that C=-50, and the unique solution to our equation is

x(t)=100-50 e^{-0.02 t}.

Observe that x increases and approaches the ratio of salt to water in the input stream as time increases. That is, there is one pound of salt in each gallon of brine and this leads, eventually, to 100 pounds of salt in 100 gallons of brine.

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