91Ó°ÊÓ

Brine is essentially a solution composed of salt dissolved in water. In this exercise, we are dealing with a specific brine concentration of 2 pounds of salt per gallon. When discussing brine concentration in differential equations, understanding the inflow of this concentration into a system is key.
Given that brine is being added to a tank at a specified rate, knowing the concentration helps determine the rate at which the mass (or amount) of salt enters the tank.
In our problem, the inflow rate of brine is 5 gallons per minute, multiplying by the concentration, we find that 10 pounds of salt are entering the tank each minute, or mathematically: - Inflow rate of salt: `5 * 2 = 10` pounds/minute.
This is crucial when forming the differential equation that governs how the salt behaves in the system over time.

Initial conditions play a significant role in solving differential equations. They provide a specific starting point that allows us to solve for any constant of integration during solutions. In this exercise, the initial condition is the amount of salt in the tank at time zero.
Since the tank initially contains pure water, it means there is no salt present at the beginning, hence: - Initial condition: `A(0) = 0` pounds of salt.
This initial state is instrumental for finding out the particular solution for the differential equation. By substituting this condition into the general solution, we can solve for constants, thereby tailoring the solution to match the physical scenario presented in the problem.

Separation of variables is a technique employed to solve differential equations, especially when they can be rearranged into a form where each side of the equation contains only one of the two variables involved.
For our exercise, the differential equation is: - \(\frac{dA}{dt} = 10 - \frac{A(t)}{100}\)
To separate variables, we rearrange terms such that all the \(A\)-related terms are on one side of the equation and the \(t\)-related terms on the other. It appears like this: - \( \frac{dA}{10 - \frac{A}{100}} = dt\)
Integrating both sides allows us to find a general solution. This method is handy when dealing with well-mixed solutions, like our brine and water, where we can distinctly separate incoming and outgoing rates with respect to time and the changing concentration.

The concept of exponential growth and decay describes how quantities grow or shrink at rates proportional to their current value. This is a common occurrence in natural and engineered systems where changes happen continuously over time.
In our problem, once the differential equation is separated and integrated, we arrive at a solution that resembles an exponential decay form: - \( A(t) = 1000(1 - e^{-t/100}) \)
The constant \(e\) is the base of the natural logarithm, and its exponent's coefficients determine the rate of growth or decay. - Here, \(100\) is a significant factor in the rate of decay, reflecting how the salt concentration approaches an equilibrium.
Understanding this process shows how the balance between the entering brine concentration and outgoing mixture leads to a gradual increase in salt, eventually stabilizing as depicted by the exponential function.