Question

solve using differential equations A tank contains 100 gallons of brine made by dissolving \( 80 \mathrm{lb} \) of salt in water. Pure water runs into the tank at a rate of 4 gallons per minute and a well stirred mixture pours out of t

Answer 1

To solve the problem, we can represent the rate of change of salt in the tank using a differential equation. By applying separation of variables and integrating, we can find the equation that represents the amount of salt in the tank at any given time.

Step-by-step explanation:

To solve this problem, we can use a differential equation to represent the rate of change of salt in the tank. Let's denote the amount of salt in the tank at time t as y(t). The rate at which salt is entering the tank is 0 lb/min since only pure water is running in. The rate at which salt is leaving the tank is the concentration of salt in the tank (y(t)/100) times the rate of flow of the mixture out of the tank, which is 4 gallons/min. Therefore, we can write the differential equation as:

dy/dt = -4y/100

Applying separation of variables and integrating, we get:

dy/y = -0.04 dt

Integrating both sides, we obtain:

ln|y| = -0.04t + C

where C is the constant of integration.

To find the particular solution, we can use an initial condition. When t = 0, y = 80 lb since the initial amount of salt in the tank is 80 lb. Substituting these values into the equation, we get:

ln|80| = C

C = ln(80)

Therefore, the equation becomes:

ln|y| = -0.04t + ln(80)

Taking the exponential of both sides, we get:

|y| = e^(ln(80)) * e^(-0.04t)

|y| = 80 * e^(-0.04t)

Since the amount of salt cannot be negative, we can remove the absolute value signs:

y = 80 * e^(-0.04t)

This equation represents the amount of salt in the tank at any given time.