91Ó°ÊÓ

The concept of "rate of change" is fundamental when working with differential equations. It describes how much a quantity, such as the amount of salt in a tank, changes over a period of time. In the context of the provided exercise, the rate at which salt is changing in the tank is determined by both its inflow and outflow rates.

Here’s a breakdown of how these rates function:

• **Inflow Rate:** Salt is entering the tank at 5 pounds per gallon, with a flow of 20 gallons per minute. Thus, the inflow rate is calculated as: \(5 \times 20 = 100 \text{ lb/min}\).
• **Outflow Rate:** The salt mixture is draining out at the same rate (20 gallons per minute). Given that there are \(y(t)\) pounds of salt in the mixture, the outflow rate becomes: \( \frac{y(t)}{10} \text{ lb/min}\).

The differential equation \( \frac{dy}{dt} = 100 - \frac{y}{10} \) defines the rate of change of salt within the tank, emphasizing that it is the inflow minus the outflow. Understanding this rate is crucial for solving when and how the amount of salt changes over time.

When faced with a linear first-order differential equation, like \( \frac{dy}{dt} + \frac{y}{10} = 100 \), employing the integrating factor method can significantly simplify finding the solution.

The integrating factor, derived from the standard form of the differential equation, is \( e^{t/10} \). Follow these steps to apply it:

• **Multiply Through by Integrating Factor:** Multiply every term in the equation by \( e^{t/10} \) to transform the left-hand side into an exact derivative. This results in: \( e^{t/10} \frac{dy}{dt} + \frac{y}{10}e^{t/10} = 100e^{t/10} \).
• **Simplify to Obtain an Exact Derivative:** The equation simplifies to: \( \frac{d}{dt}(y \, e^{t/10}) = 100e^{t/10} \). This form makes integration straightforward.
• **Integrate Both Sides:** Integrate to find \( y \, e^{t/10} = 1000e^{t/10} + C \), where \( C \) is a constant of integration.

Using the integrating factor helps convert the differential equation into a form that's much easier to integrate and solve, providing a direct path to the solution.

Initial conditions in differential equations are values of the variables at a specific point in time, which provide a starting point for the solution. They are crucial for determining the particular solution to a differential equation. In the tank problem, the initial condition given is that there is no salt in the tank at time \( t = 0 \).

This translates to the condition \( y(0) = 0 \). By applying this specific condition to the solution derived from the equation, you can solve for the constant \( C \) involved in the integration process:

• Substitute the initial condition into the integrated form: \( 0 = 1000 + C \).
• Solve for \( C \): Here, you find \( C = -1000 \).

Applying the initial condition allows us to refine our general solution \( y \, e^{t/10} = 1000e^{t/10} - 1000 \) to a particular solution that fits the specific situation of the tank initially containing no salt. This ensures the solution accurately models the real-world scenario described in the problem.