91Ó°ÊÓ

Differential equations are mathematical equations that relate a function with its derivatives. They are essential as they describe many physical and natural phenomena. In this context, the differential equation given is \( y' = y(1-y) \), a logistic growth model. Logistic growth represents a population growth pattern where growth rates slow as the population nears a maximum limit, called the carrying capacity. This S-shaped curve indicates rapid growth followed by a plateau.
The equation \( y' = y(1-y) \) specifically falls under first-order differential equations, where \( y' \) defines the rate of change of \( y \) with respect to time \( t \). Understanding this equation involves recognizing how the rate of growth relies on both the current state \( y \) and how close \( y \) is to the carrying capacity of 1. As **\( y \) approaches 1**, the value **\((1-y)\)** approaches 0, slowing the overall growth.

An initial condition is a value that helps determine an exact solution to a differential equation. It specifies the value of the function at a particular point, often at the start of an observation \((t=0)\). In our exercise, the initial condition is given as \( y(0) = \frac{1}{2} \), meaning that at time \( t=0 \), the population is half of the carrying capacity.
Applying an initial condition is crucial for finding the particular solution of a differential equation. It involves adjusting the general solution by solving for any arbitrary constants. For our logistic growth equation, this was used to determine the constant \( C \) in the expression for \( y \). Without such conditions, many potential solutions could fit the differential equation, but only one particular solution will satisfy both the differential equation and the initial condition.

Separation of variables is a technique used to solve certain differential equations. It involves rearranging the equation so that all terms in one variable (and its derivatives) are on one side and all terms in the other variable are on the opposite side.
For the logistic growth equation \( y' = y(1-y) \), the separation of variables allows us to rearrange it into the form \( \frac{dy}{y(1-y)} = dt \). Once separated, we integrate both sides to find an expression for \( y \) concerning \( t \).

• Step 1: Use partial fraction decomposition to break down \( \frac{1}{y(1-y)} \) into simpler fractions: \( \frac{1}{y} + \frac{1}{1-y} \).
• Step 2: Integrate these expressions individually over \( dy \) and find \( \int dt \).
• Step 3: Solve the resulting expression to get back to \( y \) as a function of \( t \).

This method yields a more manageable integration problem, facilitating the solution derivation. It efficiently exploits the structure of the logistic growth equation to produce the general solution that matches the initial condition.