Problem 31 - Thomas Calculus

Question

In Exercises 31 and \(32,\) obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), \quad y(0)=1 / 2} \\ {\quad 0 \leq x \leq 4, \quad 0 \leq y \leq 3}\end{array} $$

Step-by-Step Solution

Answer

Particular solution: \( y(x) = \frac{2}{1 + 3e^{-2x}} \).

Step 1: Understand the Equation Type

The given differential equation is a logistic equation of the form \( y' = y(2-y) \). Logistic equations typically describe scenarios where growth is self-limiting, such as populations growing within a finite resource environment.

Step 2: Identify the Initial Condition

The initial condition provided is \( y(0) = \frac{1}{2} \). This means that at \( x = 0 \), the value of \( y \) is \( \frac{1}{2} \). We'll use this to determine the particular solution after finding the general solution.

Step 3: Determine the General Solution Using CAS

Using a CAS (Computer Algebra System) or solving analytically, integrate the differential equation \( \frac{dy}{y(2-y)} = dx \). The logistic equation's general solution is similar to:\[y(x) = \frac{2}{1 + Ce^{-2x}}.\]

Step 4: Apply Initial Condition to Find Particular Solution

Substitute the initial condition \( y(0) = \frac{1}{2} \) into the general solution equation to solve for the constant \( C \):\[\frac{1}{2} = \frac{2}{1 + Ce^{0}} \rightarrow 1 + C = 4 \rightarrow C = 3.\]This gives the particular solution:\[y(x) = \frac{2}{1 + 3e^{-2x}}.\]

Step 5: Sketch the Slope Field

A slope field, or direction field, is a graphical representation of the differential equation showing the slope at various points. Using software or graphical tools, plot the field of small line segments at grid points, representing the slope \( y'(x) = y(2-y) \) within the specified bounds \( 0 \leq x \leq 4 \) and \( 0 \leq y \leq 3 \).

Step 6: Graph the Particular Solution

On the same axes as the slope field, plot the curve of the particular solution \( y(x) = \frac{2}{1 + 3e^{-2x}} \). Ensure that the curve passes through the initial condition point \( (0, \frac{1}{2}) \) and follows the direction suggested by the slope field.

Key Concepts

Slope Field

A slope field, or direction field, is an invaluable tool in visualizing differential equations, like our logistic equation. Imagine it as a map of arrows, where each arrow indicates the slope (or direction) that solutions would follow at any given point. For our logistic equation \( y' = y(2-y) \), the slope at each point \((x, y)\) is determined by substituting \(y\) into the equation.
To create a slope field:

• Pick several points within the given interval; in this case, \( 0 \leq x \leq 4 \) and \( 0 \leq y \leq 3 \).
• Calculate the slope at each point using the equation \( y' = y(2-y) \).
• Draw short lines or arrows at each point to represent the calculated slopes.

This visualization helps understand how solutions behave over time and can guide the drawing of particular solutions by showing the general flow dictated by the equation.

Initial Condition

The initial condition acts like a starting point for solving differential equations. In our case, it's given as \( y(0) = \frac{1}{2} \). This means simply that at \( x = 0 \), the value of \( y \) is \( \frac{1}{2} \).
Why is this important?

• It provides a specific point through which the solution must pass.
• Using it helps to find a unique solution from the general solution.

For logistic equations, this initial condition helps pin down the particular solution which acknowledges the real-world constraints at the starting point, like population at a specific initial time.

General Solution

The general solution of a differential equation represents a family of functions that satisfy the equation. For the logistic equation \( y' = y(2-y) \), integrating the equation gives us a solution of the form:\[ y(x) = \frac{2}{1 + Ce^{-2x}} \]Here, \( C \) is a constant that varies, creating different curves within this family, each representing a different specific scenario.
Finding the general solution involves:

• Separating variables to isolate terms involving \( y \) and \( x \).
• Integrating each side to find \( y \) as a function of \( x \) and the constant \( C \).
• The constant \( C \) will be determined when a particular solution is sought using a specific initial condition.

The general solution provides broad insight into how solutions relate to each other under varying conditions.

Particular Solution

A particular solution is obtained from the general solution when the initial condition is applied. For our equation, \[ y(x) = \frac{2}{1 + Ce^{-2x}} \]substitute \( y(0) = \frac{1}{2} \) to solve for \( C \):\[ \frac{1}{2} = \frac{2}{1 + C} \]This simplifies to \( C = 3 \), providing the particular solution:\[ y(x) = \frac{2}{1 + 3e^{-2x}} \] How is this result significant?

• It uniquely specifies a solution that fits the initial condition.
• In contexts like growth models, it can predict future behavior based on initial conditions.

Plotting this solution alongside the slope field shows how closely it follows the direction indicated by the slope arrows, validating the consistency of our analysis.