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Hello, someone please find the solution to this differential equation problem. P

ID: 2963881 • Letter: H

Question

Hello, someone please find the solution to this differential equation problem. Please, solve on a SEPARATE SHEET of paper. Explain your solution! Thank you :)

A tank initially contains 60 gallons of pure water. Brine containing 1 lb of stilt per gallon enters the tank at 3 gallons per minute, and the perfectly mixed solution leaves the tank at 4 gallons per minute. Find a formula for the amount of salt in the tank after t minutes. What will be the concentration of salt at the instant the tank becomes empty? That is, what will be the concentration of salt in the last few drops of water that leave the tank?

Explanation / Answer

S(t) = amount of salt at time t, S(0) = 0

Salt flows in at rate of:
1 lbs/gal * 3 gal/min= 3 lbs/min

Salt flows out at rate of:
Concentration of salt in tank * 4 gal/min = S lbs/60 gal * 4 gal/min = (S/10) lbs/min

dS/dt = 3 ? S/15
dS/dt = (45 ? S) / 15
dS/(45 ? S) = 1/15 dt

Integrate both sides:
? dS/(45 ? S) = ? 1/15 dt
? ln|45 ? S| = 0.0666t + C?
ln|45 ? S| = ?0.0666t + C?
45 ? S = e^(?0.0666t + C?) = e^(?0.0666t) * e^C?
45 ? S = C e^(?0.0666t)

When t = 0, S = 0
45 ? 0 = C e^0
C = 45

45 ? S = 45 e^(?0.0666t)
S = 45 ? 45 e^(?0.0666t)
S = 45 (1 ? e^(?0.0666t)

b)
since brine is coming at 3 gallons per min and leaving at 4 gallons per min
effectively it is going at 1 gallons per min

time taken to empty tank=60 min

putting t=60 in last equation:
s=45(1 ? e^(?0.0666*60)
=45(1 ? e^-4)
=44.175