Chapter 8: Problem 64
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). \(\frac{d N}{d t}=(N-1)(N+1)(N-4)\) (a) \(N(0)=0\), (b) \(N(0)=2\), (c) \(N(0)=6\), (d) \(N(0)=-2\).
Short Answer
Step by step solution
Understand the Differential Equation
Find Equilibrium Points
Analyze Stability at Equilibrium Points
Sketch the Vector Field
Sketch the Solutions for Initial Conditions
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field
For example, consider the differential equation \( \frac{d N}{d t} = (N-1)(N+1)(N-4) \). To draw the vector field for this equation, observe separate regions divided by equilibrium points. Before plotting, determine where the derivative is positive or negative across these regions.
When \( \frac{d N}{d t} > 0 \), draw arrows pointing to the right (indicating increasing \( N \)), and when \( \frac{d N}{d t} < 0 \), draw arrows pointing to the left (indicating decreasing \( N \)). This method helps predict how \( N \) behaves depending on its initial state.
Equilibrium Points
Find them by setting the derivative equal to zero:
- \( N = 1 \)
- \( N = -1 \)
- \( N = 4 \)
Each value represents an equilibrium point. At these points, \( N \) does not change, acting as a tipping or balancing point in the system, akin to a seesaw. They can be likened to a pause in the evolution of a system's behavior. Understanding such points is crucial for predicting system dynamics and eventual outcomes.
Stability Analysis
Evaluate the signs of the changes in \( N \) around each equilibrium point to deduce stability:
- At \( N = -1 \), if moving away from -1 in either direction decreases the function's value (\( \frac{d N}{d t} > 0 \)), it indicates stability.
- At \( N = 1 \), if a minor shift leads \( N \) in both directions to move away from that point (\( \frac{d N}{d t} > 0 \) or \( < 0 \)), it is unstable.
- At \( N = 4 \), if moving away leads to \( N \) increasing but coming back to the point is challenging (\( \frac{d N}{d t} < 0 \)), it indicates instability.
Stability analysis provides insight into how external influences impact the steady state of a system and whether eventual return to equilibrium is feasible.
Initial Conditions
Take, for example, our differential equation: with specified conditions given by
- \( N(0) = 0 \)
- \( N(0) = 2 \)
- \( N(0) = 6 \)
- \( N(0) = -2 \)
Initially,
- For \( N(0) = 0 \), \( N \) tends towards -1.
- From \( N(0) = 2 \), \( N \) trends toward 4.
- Starting with \( N(0) = 6 \), \( N \) goes off to infinity.
- And \( N(0) = -2 \) leads \( N \) further negative.