Question

a) Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds ofsalt have been dissolved. Pure water is pumped into the tank at a rate of 3gal/min, and whenthe solution is well stirred, it is then pumped out at the same rate. Determine a differentialequation for the amount of salt, , in the tank at time .

Answer 1

x = 50*e∧ -t/100

Explanation:

We assume:

1.-That the volume of mixing is always constant 300 gallons

2.-The mixing is instantaneous

Δ(x)t = Amount in - Amount out

Amount = rate * concentration*Δt

Amount in = 3 gallons/ min * 0 = 0

Amount out = 3 gallons/min * x/ 300*Δt

Then

Δ(x)t/Δt = - 3*x/300 Δt⇒0 lim Δ(x)t/Δt = dx/dt

dx/dt = - x/100

dx/ x = - dt/100

A linear first degree differential equation

∫ dx/x = ∫ - dt/100

Ln x = - t/100 + C

initial conditions to determine C

t= 0 x = 50 pounds

Ln (50) = 0/100 * C

C = ln (50)

Then final solution is:

Ln x = - t/100 + Ln(50) or

e∧ Lnx = e ∧ ( -t/100 + Ln(50))

x = e∧ ( -t/100) * e∧Ln(50)

x = e∧ ( -t/100) * 50

x = 50*e∧ -t/100