Suppose the total amount of salts in the tank is

s[0] = 50 pounds

The rate of increase of s[t] is

(3 gal/min)*(2 lb/gal) = 6 lb/min

The rate of decrease of s[t] (in pounds per minute) is

s/(300 gal)*(3 gal/min)

Expressing these relationships using a differential equation, we have

s'[t] = 6 - (1/100)*s,

s[0] = 50

We recognize that the solution to this linear non-homogeneous equation with constant coefficients will be of the form

Substituting this into the differential equation above, and matching coefficients, we get

c

s[50] = 600 - 550*e^(-1/2)

**s[t]**pounds, where t is time in minutes. We are told thats[0] = 50 pounds

The rate of increase of s[t] is

(3 gal/min)*(2 lb/gal) = 6 lb/min

The rate of decrease of s[t] (in pounds per minute) is

s/(300 gal)*(3 gal/min)

Expressing these relationships using a differential equation, we have

s'[t] = 6 - (1/100)*s,

s[0] = 50

We recognize that the solution to this linear non-homogeneous equation with constant coefficients will be of the form

**s[t] = c**_{1}**e^(a*t) + c**_{0}for some values of a and constants c_{n}.Substituting this into the differential equation above, and matching coefficients, we get

c

_{1}*a*e^(a*t) = 6 - (1/100)*c_{1}*e^(a^t) - (1/100)*c_{0}c_{1}*a = -(1/100)*c_{1}(match coefficients of e^(a*t))**a = -1/100**0 = 6 - (1/100)*c_{0}(match constants)**c**_{0}**= 600**c_{1}*e^(-0/100) + 600 = 50 (evaluate for initial condition)**c**_{1}= 50 - 600**= -550**The salt quantity is**s[t] = 600 - 550e^(-t/100))**After 50 minutes, the salt quantity in the tank will bes[50] = 600 - 550*e^(-1/2)

**s[50] ≈ 266.4 pounds of salts**The quantity present as t gets very large will be**s[infinity] = 600 pounds**