A 19000 cubic foot room initially has a radon level of 850 picocuries per cubic foot. A ventilation system is installed that brings in 500 cubic feet of air per hour that contains 4 picocuries per cubic foot, while an equal quantity of the well-mixed air in the room leaves the room each hour. Setup and use a differential equation to determine how long it will take for the room to reach a safe to breathe level of 120 picocuries per cubic foot. (Round your answer to 5 decimal places.)

As the ventilation system installed brings in 500 cubic feet of air per hour that contains 4 picocuries per cubic foot, every hour there are 500*4 = 2000 picocuries entering the room.

At the same time, 500y(t) picocuries are leaving the room.

So, the rate of change of the density of the picocuries is

(amount of picocuries at time t)/19000 =(amount of picocuries entering - amount of picocuries leaving at time t)/19000

That is to say,

[tex]y'(t)=\frac{20000-500y(t)}{19000}=\frac{200-5y(t)}{190}[/tex]

Let's write the equation in the standard mode by operating on this expression:

[tex]y'(t)+\frac{1}{38}y(t)=\frac{20}{19}[/tex]

This an ordinary linear differential equation of 1st order, whose integrating factor is

[tex]e^{t/38}[/tex]

So, the solution is

[tex]y(t)=ce^{-t/38}+40[/tex]

To find c, we use the initial value y(0)=850 and we get

c = 810

and the equation that gives the density of picocuries per cubic foot is

[tex]\boxed{y(t)=810e^{-t/38}+40}[/tex]

Now, we must look for a value of t for which y(t)=120.

[tex]y(t)=120 \Rightarrow 810e^{-t/38}+40=120\Rightarrow e^{-t/38}=\frac{8}{81}[/tex]

Taking logarithm on both sides

[tex]\frac{-t}{38}=\ln(8/81)\Rightarrow \boxed{t=87.97028\;hours}[/tex]