91Ó°ÊÓ

First-order linear differential equations are a fundamental part of calculus and are crucial in modeling various situations, like population growth, cooling of an object, or, as in our case, remembering people over time. These equations involve an unknown function, often denoted as \( y \), and its derivative \( y' \). The general form is:

• \( y' + P(t)y = Q(t) \)

This means the equation consists of the derivative of \( y \), a term involving \( y \) itself, and a function of \( t \). In the example provided, the function is \( y' + 0.2y = 30 \) which illustrates the rate of change in the number of people remembered, balancing meeting 30 new people a year and forgetting 20% of all known people.
Understanding this structure helps us recognize the problem type and choose the right solution method, such as the integrating factor method.

The integrating factor is a powerful technique used for solving first-order linear differential equations. It simplifies these equations into a form that can be easily integrated. In our context, the integrating factor \( \mu(t) \) is calculated as follows:

• \( \mu(t) = e^{\int P(t) \ dt} \)

For the differential equation \( y' + 0.2y = 30 \), the integrating factor becomes \( \mu(t) = e^{0.2t} \).

When you multiply the entire differential equation by this integrating factor, the equation can be restructured into a form where the left-hand side becomes the derivative of a product of \( y(t) \) and the integrating factor itself. This transformation is crucial because it allows us to integrate both sides easily. In essence, it unlocks a path to finding \( y(t) \), the function we aim to solve for.

Initial conditions are essential for finding specific solutions to differential equations, tailoring the general solution to the problem's context. In this exercise, the initial condition given is \( y(0) = 0 \), which indicates that at time \( t = 0 \), no people are remembered. Using initial conditions involves substituting specific values into the equation to find constants (like \( C \) in our solution).

In the example, after finding the general solution \( y(t) = 150 + Ce^{-0.2t} \), we use \( y(0) = 0 \) to determine \( C \). By substituting \( t = 0 \) into the equation, we solve \( 0 = 150 + C \cdot 1 \) and find \( C = -150 \).

• This step is crucial; it ensures the solution not only fits the mathematical model but also aligns with real-world initial conditions, thus providing a complete, accurate solution for the problem.