A tank originally holds 100 gal of water with 3 lb of salt dissolved. Then water containing LaTeX: / frac{1}{3}1 3lb of salt per gallon is poured into the tank at a rate of 5 gal/min. Well-mixed solution is leaving the tank at a rate of 1 gal/min.

A. Find the initial-value problem that describes the amount of salt in the tank? B. Find the amount of salt in the tank after time T. C. If you have access to a computer with graphing capabilities, sketch the graph of solution foind in (b).

Let S (t) denote the amount of salt in the tank at time t. Then S (0) = 3 lb.

Salt flows into the tank at a rate of

(1/3 lb/gal) * (5 gal/min) = 5/3 lb/min

and flows out at a rate of

(S (t) / V (t) lb/gal) * (1 gal/min)

where V (t) is the volume of liquid in the tank at time t. The tank starts off with 100 gal of liquid, and each min it gains a net volume of 5 - 1/3 = 14/3 gal of liquid, so that

V (t) = 100 + 14/3 t

Then the net rate of change of the amount of salt in the tank is governed by the linear differential equation,

dS/dt = 5/3 - S / (100 + 14/3 t)

or

dS/dt + 3S / (300 + 14t) = 5/3

To solve this ODE, multiply both sides by the integrating factor (300 + 14t) ^ (3/14), so that the left side can be condensed to the derivative of a product:

(300 + 14t) ^ (3/14) dS/dt + 3 (300 + 14t) ^ (-11/14) S = 5/3 (300 + 14t) ^ (3/14)

d/dt [ (300+14t) ^ (13/14) S] = 5/3 (300 + 14t) ^ (3/14)

Integrate both sides with respect to t, then solve for S:

(300+14t) ^ (13/14) S = 5/51 (300 + 14t) ^ (17/14) + C

S = 5/51 (300 + 14t) ^ (4/14) + C (300+14t) ^ (-13/14)

Given that S (0) = 3, we find

3 = 5/51 * 300^ (4/14) + C * 300^ (-13/14)

==> C ≈ 498.99

Then the amount of salt in the tank at time t is given by

S (t) ≈ 5/51 (300 + 14t) ^ (4/14) + 498.99 (300+14t) ^ (-13/14)