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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.R.19d
Chapter 9, Problem 9.R.19d

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.
d. For what values of A are the corresponding solutions decreasing, for t≥0
Direction field graph showing slope vectors for y′=y(2−y) with axes labeled t and y from -3 to 3.

Verified step by step guidance
1
Identify the differential equation given: \(y' = y(2 - y)\). This represents the slope of the solution curves at any point \((t, y)\).
Recall that the solution is decreasing where the derivative \(y'\) is negative. So, we need to find where \(y(2 - y) < 0\).
Analyze the inequality \(y(2 - y) < 0\). This product is negative when one factor is positive and the other is negative. So, either \(y < 0\) and \(2 - y > 0\), or \(y > 0\) and \(2 - y < 0\).
Simplify the conditions: For \(y < 0\), \(2 - y\) is always positive, so \(y' < 0\) for all \(y < 0\). For \(y > 0\), \(2 - y < 0\) means \(y > 2\). So, \(y' < 0\) when \(y > 2\).
Conclude that the solutions are decreasing for \(t \geq 0\) if the initial condition \(A = y(0)\) satisfies \(A < 0\) or \(A > 2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Direction Fields and Slope Interpretation

A direction field visualizes the slopes of solutions to a differential equation at various points. Each small line segment represents the slope y' at that (t, y) coordinate, indicating the behavior of solution curves without solving the equation explicitly.
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Understanding Slope Fields

Equilibrium Solutions and Stability

Equilibrium solutions occur where y' = 0, meaning the solution is constant. For y' = y(2 - y), equilibria are at y = 0 and y = 2. Stability depends on the sign of y' near these points, determining if solutions approach or move away from equilibria.
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Solutions to Basic Differential Equations

Monotonicity of Solutions Based on Initial Conditions

The sign of y' determines whether solutions increase or decrease. For initial value y(0) = A, if y' < 0 for t ≥ 0, the solution decreases. Analyzing y' = y(2 - y) shows that solutions decrease when A > 2 or A < 0, as the product y(2 - y) is negative in these intervals.
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Initial Value Problems
Related Practice
Textbook Question

A second-order equation Consider the equation


t² y′′(t) + 2ty′(t) − 12y(t) = 0


b. Assuming the general solution of the equation is


y(t) = C₁ tᵖ¹ + C₂ tᵖ²,


find the solution that satisfies the conditions


y(1) = 0, y′(1) = 7

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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.

e. Discuss some possible shortcomings of this model. Why might the carrying capacity be either greater than or less than the value predicted by the model?

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Textbook Question

11–18. Solving initial value problems Use the method of your choice to find the solution of the following initial value problems.

y′(x) = x/y, y(2) = 4

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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) = (2t+1)(y²+1)

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Textbook Question

Logistic growth The population of a rabbit community is governed by the initial value problem

P′(t) = 0.2 P (1 − P/1200), P(0) = 50

d. What is the population when the growth rate is a maximum?

Textbook Question

A first-order equation Consider the equation t² y′(t) + 2ty(t) = e⁻ᵗ

c. Find the solution that satisfies the condition y(1) = 0