Question

draw a direction field and plot (or sketch) several solutions of the given differential equation. Describe how solutions appear to behave as \(t\) increases, and how their behavior depends on the initial value \(y_{0}\) when \(t=0\). $$ y^{\prime}=t y(3-y) $$

Short answer

Based on the given differential equation \(y^{\prime}=ty(3-y)\), describe how the solutions behave as time increases and depending on the initial value. As time increases, the solutions behave as follows: 1. When the initial value \(y_{0}=0\), the solution stays at \(y=0\) for all t. 2. When the initial value \(y_{0}>0\) and \(y_{0}<3\), the solution increases but with a diminishing rate as it approaches \(y=3\). 3. When the initial value \(y_{0}=3\), the solution stays at \(y=3\) for all t. 4. When the initial value \(y_{0}>3\), the solution decreases but with a diminishing rate as it approaches \(y=3\). As time increases, it appears that all solutions are attracted to the line \(y=3\) and likely tend towards this value.

Step-by-step solution

Find the equilibrium points

To find where \(y^{\prime}=0\), set the right-hand side of the equation equal to zero: $$ty(3-y)=0$$ The equilibrium points (where \(y^{\prime}=0\)) are when \(t=0\) or \(y=0\) or \(y=3\). We will use these equilibrium points to help analyze the solution curves of the direction field.

Create the direction field

To create the direction field, we will evaluate the differential equation, \(y^{\prime}=ty(3-y)\), at several points in the t-y plane. Since there are no explicit t or y terms within the right-hand side of the equation, the direction field only depends on t and y. Using equilibrium points, we know that the slope of the solutions equal 0 along the lines t=0, y=0, and y=3. Based on this information, we can sketch the direction field on a t-y plane.

Sketch the solutions

Using the direction field, we can now sketch some solutions. Select different initial conditions and follow the direction field to draw solution curves. Note that the solutions will be horizontal lines when \(y=0\) or \(y=3\), and vertical lines when \(t=0\). Since the solutions will lie upon the direction field, we can draw the solutions by looking at how they interact with the lines mentioned in the previous step.

Analyze the behavior of the solutions as t increases

Based on the sketches we have made, we can describe how the solutions behave as time increases: 1. When \(y_{0}=0\), the solution stays at \(y=0\) for all t. 2. When \(y_{0}>0\) and \(y_{0}<3\), the solution increases but with a diminishing rate as it approaches \(y=3\). 3. When \(y_{0}=3\), the solution stays at \(y=3\) for all t. 4. When \(y_{0}>3\), the solution decreases but with a diminishing rate as it approaches \(y=3\). As \(t\) increases, it appears that all solutions are attracted to the line \(y=3\) and likely tend towards this value.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental tools for describing and analyzing how a system changes with respect to time or space.
In the given problem, we have a differential equation of the form \( y' = ty(3-y) \). This tells us how the rate of change of \( y \) with respect to \( t \) is influenced by both \( t \) and \( y \).

• \( y' \) represents the derivative of \( y \) with respect to \( t \).
• \( ty(3-y) \) shows that the rate of change depends on the current values of \( t \) and \( y \), highlighting a dynamic system.

Differential equations like this one can model a variety of real-world systems, such as population growth or the spread of diseases.

Equilibrium Points

Equilibrium points are values where the system does not change. In other words, these are the points where the derivative \( y' \) equals zero, leading to a stable state.
For the equation \( ty(3-y)=0 \), equilibrium points occur when \( t=0 \), \( y=0 \), and \( y=3 \).

• At \( t = 0 \), the time or horizontal axis interacts with the plane, offering potential equilibrium due to time intervention alone.
• \( y = 0 \) and \( y = 3 \) are equilibrium states due to the fact that when \( y \) either matches 0 or 3, one of the factors becomes zero, causing \( y' \) to be zero.

The concept of equilibrium is crucial since it helps in understanding the stability and long-term behavior of a solution in a dynamical system.

Solution Curves

Solution curves are graphs that show all possible solutions to a differential equation for various initial conditions. They help visualize how a system evolves over time or space.
When you sketch solutions from the direction field, you choose different starting points and trace paths that follow the slopes suggested by the direction field. These paths are your solution curves.

• In our context, solution curves will be horizontal at \( y = 0 \) and \( y = 3 \) since \( y' = 0 \) in these regions.
• Vertical lines will be positioned along \( t = 0 \), showing no directional slope change at this point in time.
• Each curve reflects a unique evolution of the solution based on its starting value \( y_0 \).

Solution curves provide insight into what happens as \( t \) increases and help us see how the system evolves over time. They're like roadmaps of the system's potential paths.

Initial Value Problem

An initial value problem involves solving a differential equation with a given starting point or initial condition. This provides a specific path or solution within all possible solution curves.
The initial value \( y_0 \) is critical because it dictates how the solution curve starts and subsequently behaves. For example, consider these scenarios:

• If \( y_0=0 \), the solution will remain flat along \( y=0 \) for all future \( t \).
• For \( 0 < y_0 < 3 \), the solution initially increases, approaching \( y=3 \) as \( t \) increases, yet never crossing beyond it.
• With \( y_0=3 \), the solution remains steady at \( y=3 \).
• For \( y_0 > 3 \), the solution decreases towards \( y=3 \), much like a damped oscillation, though in a monotonic approach.

Understanding the initial value problem is important for predicting the system's future behavior from any given starting point. It also suggests the system's long-term trends or asymptotic states.