Differential Equations

Question

A 100 L tank contains 1 L of a brine solution with 2 kg of salt dissolved in it. At time t = 0, another brine solution flows into the tank at a rate of 6 L/min, however, this brine solution contains a variable concentration of salt; in particular, it contains 1/(1+t²) kg/L of salt at time t for 0 ≤ t≤ 33. At the same time, a valve is opened at the bottom of the tank, allowing the (well-mixed) solution to flow out at a rate of 3 L/min. Find a formula for the amount of salt in the tank at time t for 0 ≤ t ≤ 33.

Answer 1

Interesting question.

x(t) = amount of salt in the tank, in kg (this is not the concentration but the mass).

V(t) = volume of liquid in the tank, in Liters = 1 + 6t - 3t (initial 1L and 6L/min going in, 3L/min going out)

dx/dt = 6* 1/(1+t^2) - 3 * x / V(t)

Plug in V(t) and solve the diff. equation with the initial condition x(0) = 2kg salt.