Question

A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution

Answer 1

Set up a differential equation using rate of salt going in minus rate going out.

Rate = dS/dt = salt/gal* gal/min

Also the number of gallons in tank is increasing at (4-2) gal/min = 2 gal/min
The volume of tank at any given time = 100+2t

Differential Equation:
$\frac{dS}{dt} = (\frac{1}{4})(4) - (\frac{S}{100+2t}) (2) \\ \\ \frac{dS}{dt} = 1 - \frac{S}{t + 50} \\ \\ \frac{dS}{dt} + (\frac{1}{t+50}) S = 1$

Use integrating factor of t+50
$[S(t+50)]' = t+50$

Integrate both sides
$S(t+50) = \frac{t^2}{2} + 50t + c$
Solve for constant c given S(0) = 20
c = 1000

Tank is full at 200 gallons when t =50
100+2t = 200 ---> t = 50

S(50) = 47.5 

Final Answer: There are 47.5 pounds of salt