simple differential eq problem?

Question

Suppose that a large mixing tank initially holds 300 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 2lb/gal, determine a differential equation for the amount of salt A(t) in the tank at time t.

Answer 1

I've answered a similar question before:

http://answers.yahoo.com/question/index;_ylt=Ao08KaiA7Z_AneKbO1TRdCDty6IX?qid=20070731231233AAh9kjp&show=7#profile-info-P9yp8KNSaa

Just replace CO2 concentrations with your salt concentrations.

The only real trick here is that your volume changes with time as well.

salt in = ds/dt = (2lbs/ga)(3gal/min) - (s/V)(2gal/min)

where s is the amount of salt in the tank and V is the volume of solution in the tank,

V = (500 + t/min)gal since 1gal/min more is pumped in than pumped out.

Thus:

ds/dt = 6lbs/min - 2s/[(500 + t/min)min]

Let w = 500 + t/min, then dt = min(dw), substitute:

ds/(min(dw)) = 6lbs/min -2s/(w(min)) or

ds/dw + 2s/w = 6lbs

Multiply everything by the integrating factor (details in link above) which in this case is w²:

w²(ds/dw) + 2sw = 6lbs(w²)

note that the left side is

d/dw(w²(s)) = 6lbs(w²)

Integrate from initial w0 to final w using the fundamental theorem of calculus:

w²s -(w0)²s0 = 2lbs(w^3 - (w0)^3)

then

s = (w0/w)²s0 + 2lbs(w - w0^3/w²)

Substitute in for w to get back in terms of t and putting in initial values for s0

s = [(500)/(500 + t/min)]²(50lbs) + 2lbs[(500 + t/min) - 500^3/(500 + t/min)]

s(t=0) = 50lbs as it should . . .

The concentration as t --> infinity is

2lbs(t/min)/[(t/min)gal] = 2lbs/gal as it should, it's the concentration of stuff coming in and is independent of time.