Finding differential equation (mixing problem)

The problem states the following:
A full tank initially contains 100 litres of pure water. A brine solution containing 2 grams of salt per litre is pumped in at 3 litres per minute. The solution in the tank is well mixed and pumped out at 4 litres per minute. What is the volume of fluid in the tank after time t? Write down a differential equation for the amount of salt x in grams in the tank and verify that:
x(t) = 2(100 − t) + A(100 − t)^4

So, for the first part I got that V = 100+(3t-5t) at time t (in here I just calculated the amount of liquid as I was not sure if when they say 'fluid' will that also include salt or not)

As for the differential equation I did the following:
Let x be the amount of salt in the tank at time t, then the concentration at time t is x/100.
Now in the short period of time between delta(t) and delta(t) + t, the amount of salt entering the tank is 2*3*delta(t) = 6*delta(t)
Amount of salt leaving the tank:
(x/100)*4*delta(t) = (x/25)*delta(t)

Hence, increase in salt in the tank will be:
delta(x) = 6delta(t) - (x/25)delta(t)
and so
(dx/dt) = 6 - (x/25)
the solution to which is x = 150 - Ae^(-t/25) where A is the constant of integration

In the question they also give this expression: x(t) = 2(100 − t) + A(100 − t)^4 but I am not sure where they got it from. I thought of using Taylor expansion for the exponential but it did not seem to help as there were huge numbers in denominator and it looked nothing like the suggested expression.

Reply 1

ghostwalker

17

Original post by spacewalker

The problem states the following:
A full tank initially contains 100 litres of pure water. A brine solution containing 2 grams of salt per litre is pumped in at 3 litres per minute. The solution in the tank is well mixed and pumped out at 4 litres per minute. What is the volume of fluid in the tank after time t? Write down a differential equation for the amount of salt x in grams in the tank and verify that:
x(t) = 2(100 − t) + A(100 − t)^4

So, for the first part I got that V = 100+(3t-5t) at time t (in here I just calculated the amount of liquid as I was not sure if when they say 'fluid' will that also include salt or not)

As for the differential equation I did the following:
Let x be the amount of salt in the tank at time t, then the concentration at time t is x/100.

Not followed it all through, but the first thing that strikes me as incorrect is - in red. Volume is not 100 at time t.