ODE from Initial Value Problem

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SUMMARY

The discussion centers on solving an initial value problem involving a tank with 3000 liters of polluted water, initially at a 5% concentration. The differential equation governing the pollution level is established as y' = -50y, leading to the solution y = 150e^{-50t}. The goal is to determine the time required for the concentration to decrease to 1%, which translates to finding t when y equals 30. The calculated time of 2 seconds is deemed unreasonable, indicating a need for a more accurate model of the removal and replacement process.

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Homework Statement



A tank contains 3000 liters polluted water. The concentration of pollution is 5%. 50 liters per minute of the polluted water is removed and replaced with 50 liters of unpolluted water per minute. How much time will it take for the concentration to go down to 1%?

2. Homework Equations / 3. The Attempt at a Solution

It is obviously a differential equation as an initial value problem with differential equations.

y is a function of t and is the amount of pollutions in the tank after t minutes.

[tex]y(0) = 150[/tex]

The amount of pollutants in the water that is removed per minute is [itex]50 \cdot y(t)[/itex]

Then the rate of change of the pollutions in the tank is negative with a value of

[tex]y' = -50y[/tex]

which is the differential equation that can be solved.

[tex]y = Ce^{-50t}[/tex]

The initial value gives the constant C a value of 150, so the function becomes:

[tex]y = 150e^{-50t}[/tex]

Then I should find t when y is 30 (0.01 times 3000). Algebraic manipulation gives

[tex]\ln \frac{30}{150} = -50t[/tex]

This gives t equals to 2 seconds, which is unreasonable, but I do not know quite where to go from here.
 
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You need to carefully model the removal & replacement process and how it changes the % of polluted water over time.
 

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