Answer:
The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is  .
.
Step-by-step explanation:
We are given that a large mixing tank initially holds 500 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. 
The concentration of the solution entering is 4 lb/gal.
Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;
 
 
where,  = concentration of salt in the inflow
 = concentration of salt in the inflow  input rate of brine solution
 input rate of brine solution
and  = concentration of salt in the outflow
 = concentration of salt in the outflow  outflow rate of brine solution
 outflow rate of brine solution
So,  = 4 lb/gal
 = 4 lb/gal  3 gal/min = 12 lb/gal
 3 gal/min = 12 lb/gal
Now, the rate of accumulation = Rate of input of solution - Rate of output of solution
                                                 = 3 gal/min - 2 gal/min
                                                 = 1 gal/min.
It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.
So,  = concentration of salt in the outflow
 = concentration of salt in the outflow  outflow rate of brine solution
 outflow rate of brine solution
              =  =
 = 
Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;
=  
 
or  .
.