A tank contains 200 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A (t) of grams of salt in the tank at time t.

At the start, the tank contains A (0) = 50 g of salt.

Salt flows in at a rate of

(1 g/L) * (5 L/min) = 5 g/min

and flows out at a rate of

(A (t) / 200 g/L) * (5 L/min) = A (t) / 40 g/min

so that the amount of salt in the tank at time t changes according to

A' (t) = 5 - A (t) / 40

Solve the ODE for A (t):

A' (t) + A (t) / 40 = 5

e^ (t/40) A' (t) + e^ (t/40) / 40 A (t) = 5e^ (t/40)

(e^ (t/40) A (t)) ' = 5e^ (t/40)

e^ (t/40) A (t) = 200e^ (t/40) + C

A (t) = 200 + Ce^ (-t/40)

Given that A (0) = 50, we find

50 = 200 + C = => C = - 150

so that the amount of salt in the tank at time t is

A (t) = 200 - 150 e^ (-t/40)