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)