Question

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)(y+2)$$

Short answer

Question: Find the equilibrium solutions, sketch the direction field for \(t \geq 0\), and sketch the solution curve that corresponds to the initial condition \(y(0) = 1\) for the autonomous differential equation \(y^{\prime}(t) = y(y-3)(y+2)\).

Step-by-step solution

1. Find the equilibrium solutions

To find the equilibrium solutions, we need to set \(y^{\prime}(t) = 0\) and solve for \(y\). Given the differential equation $$y^{\prime}(t) = y(y-3)(y+2),$$ we have $$0 = y(y-3)(y+2).$$ This equation has three roots: \(y=0, y=-2,\) and \(y=3\). Therefore, there are three equilibrium solutions: \(y(t) = 0\), \(y(t) = -2\), and \(y(t) = 3\).

2. Sketch the direction field

To sketch the direction field, we need to draw arrows representing the slope of the tangent to the solution curve at various points in the \(ty\)-plane. Since our equation is autonomous, the direction field only depends on \(y\). Thus, we can analyze the signs of \(y^{\prime}(t)\) as follows: - For \(y<-2\): The function \(y^{\prime}(t) = y(y-3)(y+2)\) is negative (three negative factors); hence, the direction goes up and to the left. - For \(-23\): The function \(y^{\prime}(t) = y(y-3)(y+2)\) is positive (three positive factors); hence, the direction goes up and to the right. Using this information, sketch the direction field for \(t\geq0\) with horizontal lines at \(y=-2,0,3\) representing the equilibrium solutions.

3. Sketch the solution curve with initial condition \(y(0)=1\)

To sketch the solution curve that corresponds to the initial condition \(y(0) = 1\), locate the point \((0, 1)\) in the \(ty\)-plane. Since the initial condition falls in the region where \(0<y<3\), we know that \(y^{\prime}(t) = y(y-3)(y+2)\) is negative in this region, and the solution curve will go up and to the left. Based on the direction field, sketch a solution curve that starts at \((0, 1)\) and follows the arrows in the direction field. Make sure the curve does not cross the equilibrium solutions at \(y = -2\), \(y = 0\), and \(y = 3\).

Key concepts

Equilibrium Solutions

In an autonomous differential equation like \( y'(t) = y(y-3)(y+2) \), equilibrium solutions are found by solving \( y'(t) = 0 \). These are the values of \( y \) where the rate of change is zero, meaning the solution stays constant over time.

For our equation, solving \( y(y-3)(y+2) = 0 \) gives us three equilibrium solutions: \( y = 0 \), \( y = 3 \), and \( y = -2 \).

These represent horizontal lines in the direction field and imply that if the system starts at one of these values, it remains there indefinitely. Such solutions highlight the stability of these points in the dynamic system.

Direction Field

The direction field of a differential equation provides a visual guide for understanding how solutions behave based on initial conditions.

For autonomous equations, like the one given, the direction field depends only on the value of \( y \), not on \( t \).

### Constructing the Direction Field

• For \( y < -2 \): The expression is negative, indicating a downward slope.
• For \( -2 < y < 0 \): The expression is positive, showing an upward slope.
• For \( 0 < y < 3 \): Again negative, leading to a downward slope.
• For \( y > 3 \): It's positive, suggesting an upward slope.

By sketching these slopes, we create a pattern that helps us visualize how the solution curves move across the plane, adhering to these directional guidelines.

Solution Curve

A solution curve is a particular solution to the differential equation given an initial condition. It represents the path or trajectory within the direction field.

With an initial condition, such as \( y(0) = 1 \), we start at the point \((0, 1)\) in the \( ty \)-plane.

Since \( 0 < y < 3 \), the slope is negative, and thus the curve begins moving downward from left to right.

### Visualizing the Solution

• The curve should smoothly follow the arrows laid out by the direction field.
• Make sure not to cross any of the equilibrium lines (\( y = -2, 0, 3 \)).
• The curve expresses the trajectory of \( y \) with respect to \( t \) for the given start point.

Initial Conditions

Initial conditions specify the starting point for a solution curve in a differential equation. They are essential for determining the specific path a solution will take in the field of possible trajectories.

Given \( y(0) = 1 \), this point becomes our launch point on the \( ty \)-plane to begin drawing the solution curve.

### Importance of Initial Conditions

• The direction field provides multiple possible paths; initial conditions help select one specific trajectory.
• Changing the initial condition changes the starting point and hence the path of the solution curve.
• They ensure solutions adhere uniquely to the physical or theoretical scenario being modeled.