Question

(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.4kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. Answer the following questions.
1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t).
Answer (in kilograms per minute): dy/dt=
2. How much salt is in the tank after 50 minutes?
Answer (in kilograms):

Answer 1

1. dy/dt is the net rate of change of salt in the tank over time. As such, it's equal to the difference in the rates at which salt enters and leaves the tank.

The inflow rate is

(0.4 kg/L) (6 L/min) = 2.4 kg/min

and the outflow rate is

(concentration of salt at time t) (4 L/min)

The concentration of salt is the amount of salt (in kg) per unit volume (in L). At any time t > 0, the volume of solution in the tank is

100 L + (6 L/min - 4 L/min) t = 100 L + (2 L/min) t

That is, the tank starts with 100 L of pure water, and every minute 6 L of solution flows in and 4 L is drained, so there's a net inflow of 2 L of solution per minute. The amount of salt at time t is simply y(t). So, the outflow rate is

(y(t)/(100 + 2t) kg/L) (4 L/min) = 2 y(t) / (50 + t) kg/min

and the differential equation for this situation is

$\dfrac{dy}{dt} = 2.4 \dfrac{\rm kg}{\rm min} - \dfrac{2y}{50+t} \dfrac{\rm kg}{\rm min}$

There's no salt in the tank at the start, so y(0) = 0.

2. Solve the ODE. It's linear, so you can use the integrating factor method.

$\dfrac{dy}{dt} = 2.4 - \dfrac{2y}{50+t}$

$\dfrac{dy}{dt} + \dfrac{2}{50+t} y = 2.4$

The integrating factor is

$\mu = \displaystyle \exp\left(\int \frac{2}{50+t} \, dt\right) = \exp\left(2\ln|50+t|\right) = (50+t)^2$

Multiply both sides of the ODE by µ :

$(50+t)^2 \dfrac{dy}{dt} + 2(50+t) y = 2.4 (50+t)^2$

The left side is the derivative of a product:

$\dfrac{d}{dt}\left[(50+t)^2 y\right] = 2.4 (50+t)^2$

Integrate both sides with respect to t :

$\displaystyle \int \dfrac{d}{dt}\left[(50+t)^2 y\right] \, dt = \int 2.4 (50+t)^2 \, dt$

$\displaystyle (50+t)^2 y = \frac{2.4}3 (50+t)^3 + C$

$\displaystyle y = 0.8 (50+t) + \frac{C}{(50+t)^2}$

Use the initial condition to solve for C :

$y(0) = 0 \implies 0 = 0.8 (50+0) + \dfrac{C}{(50+0)^2} \implies C = -100,000$

Then the amount of salt in the tank at time t is given by the function

$y(t) = 0.8 (50+t) - \dfrac{10^5}{(50+t)^2}$

so that after t = 50 min, the tank contains

$y(50) = 0.8 (50+50) - \dfrac{10^5}{(50+50)^2} = \boxed{70}$

kg of salt.