Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
a. The differential equation y′+2y=t is first-order, linear, and separable.
Ch. 9 - Differential Equations
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 9, Problem 9.R.23
22–25. Equilibrium solutions Find the equilibrium solutions of the following equations and determine whether each solution is stable or unstable.
y′(t) = y(3+y)(y-5)
Verified step by step guidance1
Identify the equilibrium solutions by setting the derivative equal to zero: solve the equation \(y'(t) = y(3 + y)(y - 5) = 0\) for \(y\).
Find the values of \(y\) that satisfy \(y = 0\), \(3 + y = 0\), and \(y - 5 = 0\). These will give the equilibrium points.
Determine the stability of each equilibrium by analyzing the sign of \(y'(t)\) around each equilibrium point. This can be done by choosing test values slightly less and greater than each equilibrium and evaluating the sign of \(y'(t)\).
Alternatively, compute the derivative of the right-hand side function \(f(y) = y(3 + y)(y - 5)\) with respect to \(y\), denoted \(f'(y)\), and evaluate it at each equilibrium point. The sign of \(f'(y)\) at the equilibrium indicates stability: if \(f'(y) < 0\), the equilibrium is stable; if \(f'(y) > 0\), it is unstable.
Summarize the equilibrium points along with their stability classification based on the sign analysis or derivative test.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equilibrium Solutions
Equilibrium solutions are constant solutions to a differential equation where the derivative equals zero. For y′(t) = f(y), these occur at values of y where f(y) = 0. They represent points where the system remains steady over time.
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Stability of Equilibrium Solutions
Stability refers to whether solutions near an equilibrium point move towards or away from it over time. If small perturbations decay and solutions return to equilibrium, it is stable; if they grow away, it is unstable. This is often analyzed using the sign of the derivative of f(y) at the equilibrium.
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Sign Analysis of the Derivative Function
To determine stability, analyze the sign of f(y) = y(3+y)(y-5) around equilibrium points. By checking intervals between roots, you can see if solutions increase or decrease, indicating whether the equilibrium attracts or repels nearby solutions.
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Related Practice
Textbook Question
Direction fields Consider the direction field for the equation y′=y(2−y) shown in the figure and initial conditions of the form y(0)=A.
a. Sketch a solution on the direction field with the initial condition y(0)=1.
Textbook Question
Logistic growth parameters A cell culture has a population of 20 when a nutrient solution is added at t=0. After 20 hours, the cell population is 80 and the carrying capacity of the culture is estimated to be 1600 cells.
c. After how many hours does the population reach half of the carrying capacity
Textbook Question
2–10. General solutions Use the method of your choice to find the general solution of the following differential equations.
y′(t) + 2y = 6
Textbook Question
Logistic growth in India The population of India was 435 million in 1960 (t=0) and 487 million in 1965 (t=5). The projected population for 2050 is 1.57 billion.
b. Use the solution of the logistic equation and the 2050 projected population to determine the carrying capacity.
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Textbook Question
2–10. General solutions Use the method of your choice to find the general solution of the following differential equations.
y′(t) = √(y/t)