Suppose that a large mixing tank initially holds 400 gallons

Suppose that a large mixing tank initially holds 400 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 3 lb/gal, determine a differential equation for the amount of salt A(t) in the tank at time

t > 0.

(Use A for

A(t).)

=

lb/min

Solution

Given:

s 400 gallons of water in which 50 pounds of salt have been dissolved.

Another brine solution is pumped into the tank at a rate of 3 gal/min,

and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

If the concentration of the solution entering is 3 lb/gal,

then this is how I\'ve set up the equation:
dA/dt = (3 lb/gal)(3 gal/min) - (A(t)/400gal)(2 gal/min)

Dropping the units,

dA/dt = 9 - (A/200)
dA/dt + (1/200)A = 9

This is first order linear in A, and so has the integrating factor e^( (1/150) dt) = e^(t/150)

Multiply both sides by the integrating factor:

e^(t/200) [dA/dt + (1/200)A] = 9e^(t/200)

(d/dt)[e^(t/200) A] = 9e^(t/200)

Integrate both sides with respect to t from 0 to t and use the initial condition A(0) = 50. The left side is

e^(t/200) A(t) - e^(0/200) A(0)