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).

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)