Problem 60 - Calculus for Biology and Medicine

Question

Draw the vector field plot of the differential equation. Then, using the given initial conditions, sketch the solutions (i.e., draw a graph showing the dependent variable as a function of the independent variable). \(\frac{d y}{d t}=y^{2}-1\) (a) \(y(0)=-1\), (b) \(y(0)=-1 / 2\), (c) \(y(0)=1 / 2\), (d) \(y(0)=2\).

Step-by-Step Solution

Answer

Plot the vector field showing solutions attracted to \( y = -1 \) and repelled from \( y = 1 \). Use initial conditions to sketch each specific path.

Step 1: Analyze the Differential Equation

The differential equation given is \( \frac{d y}{d t} = y^2 - 1 \). This is an autonomous differential equation, meaning that the rate of change of \( y \) depends only on \( y \) itself and not on \( t \). This equation describes a vector field where the slope of the tangent to the solution curve at any point is \( y^2 - 1 \).

Step 2: Determine Equilibrium Points

Equilibrium points occur where \( \frac{d y}{d t} = 0 \). Thus, set \( y^2 - 1 = 0 \), which gives the solutions \( y = 1 \) and \( y = -1 \). These are equilibrium points on the vector field where the slope is zero.

Step 3: Classify the Stability of Equilibrium Points

To classify the stability, analyze the sign of \( \frac{d y}{d t} = y^2 - 1 \) around these points. For \( y > 1 \), \( \frac{d y}{d t} > 0 \); for \( y \in (-1, 1) \), \( \frac{d y}{d t} < 0 \); and for \( y < -1 \), \( \frac{d y}{d t} > 0 \). Therefore, \( y = 1 \) is unstable and \( y = -1 \) is stable since solutions are attracted to \( y = -1 \) and repelled from \( y = 1 \).

Step 4: Sketch the Vector Field

Create a plot with the \( y \)-axis showing \( y \) values and the \( t \)-axis as the independent variable. Draw small line segments indicating slope \( y^2 - 1 \) for various \( y \) values. At \( y = 1 \) and \( y = -1 \), the segments are horizontal as these are equilibrium points. For \( y > 1 \) and \( y < -1 \), draw segments pointing upward, and for \( -1 < y < 1 \), draw segments pointing downward.

Step 5: Solve for Particular Solutions with Initial Conditions

Use the initial conditions to draw the specific solution curves on the vector field plot.- For (a) \( y(0) = -1 \), the solution stays constant at \( y = -1 \).- For (b) \( y(0) = -1/2 \), the solution curve decreases toward \( y = -1 \).- For (c) \( y(0) = 1/2 \), the solution curve decreases toward \( y = -1 \).- For (d) \( y(0) = 2 \), the solution curve increases indefinitely.

Step 6: Sketch the Solutions

Draw the solution curves based on the vector field:- For (a) \( y = -1 \), it's a horizontal line.- For (b) \( y = -1/2 \), the curve moves down toward the line at \( y = -1 \).- For (c) \( y = 1/2 \), the curve similarly moves down toward \( y = -1 \).- For (d) \( y = 2 \), the curve moves upwards and away to higher values.

Key Concepts

Equilibrium Points

In the realm of autonomous differential equations, equilibrium points are crucial as they indicate where a system tends to remain constant over time. The general form of our equation, \(\frac{dy}{dt} = y^2 - 1\), showcases these points well. Equilibrium points occur where the rate of change is zero, which means no movement in the \(y\) direction. To find these, you set \(y^2 - 1 = 0\). Solving gives you \(y = 1\) and \(y = -1\).

These points are where the solution curves "flatten out," leading to horizontal tangent lines in the vector field. Think of equilibrium points as the calm spots in the sea where the water lies still. Because these are equilibrium points, the derivative \(\frac{dy}{dt}\) is exactly zero at these locations. This implies that any function passing through these points remains unchanged in value, leading to steady states in a dynamical system.

Stability Analysis

Understanding stability is essential to determine the behavior of a differential equation's solutions over time. For our equation \(\frac{dy}{dt} = y^2 - 1\), examining stability means looking at how slight deviations near these equilibrium points transform.

Let's break down their behavior:

• For \(y > 1\), the derivative is positive, meaning solutions tend to increase and move away from the equilibrium at \(y = 1\).
• Within \(-1 < y < 1\), we find the derivative to be negative. Solutions decrease, moving toward \(y = -1\), making this region asymptotically stable.
• For \(y < -1\), \(\frac{dy}{dt} > 0\) again, solutions tend to move upward and away from the \(y = -1\).

Thus, \(y = 1\) is considered an unstable point because solutions veer away from it, while \(y = -1\) is stable as solutions are attracted back to this point. Think of stability analysis as predicting how a small community behaves near a village square—either staying close-knit or dispersing based on shared dynamics.

Vector Field Plots

Vector field plots provide visual insight into the behavior of solutions to differential equations. By plotting tiny vectors and observing their direction, we gain intuitive understanding of the system dynamics.

In our plot for \(\frac{dy}{dt} = y^2 - 1\):

• Draw a vertical axis for \(y\) values and a horizontal axis for time \(t\).
• Represent each point \(y\) with a line segment indicating \(y^2 - 1\), denoting slope direction and rate.
• At the equilibrium points \(y = 1\) and \(y = -1\), lines are horizontal, as these areas are stable or unstable plateaus.
• Above and below equilibrium points, the vector's slope shows the increasing (upward) or decreasing (downward) trends based on \(\frac{dy}{dt}\).

Vector field plots act like roadmaps, showing pathways across terrains of mathematical landscapes. They provide a framework to predict how the solutions to differential equations unfold over time, offering both quantitative and qualitative insights.