Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 9 - Differential Equations

Problem 9.2.39a

Chapter 9, Problem 9.2.39a

38–43. Equilibrium solutions A differential equation of the form y′(t)=f(y) is said to be autonomous (the function f depends only on y). The constant function y=y0 is an equilibrium solution of the equation provided f(y0)=0 (because then y'(t)=0 and the solution remains constant for all t). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations.
a. Find the equilibrium solutions.

y′(t) = 6 - 2y

Verified step by step guidance

Identify the given autonomous differential equation: \(y'(t) = 6 - 2y\).

Recall that equilibrium solutions occur where the derivative \(y'(t)\) is zero, meaning the function \(y(t)\) does not change over time.

Set the right-hand side of the differential equation equal to zero to find equilibrium points: \(6 - 2y = 0\).

Solve the algebraic equation for \(y\) to find the equilibrium value(s): \(2y = 6\) which implies \(y = 3\).

Conclude that the equilibrium solution is the constant function \(y(t) = 3\), which corresponds to a horizontal line in the direction field.

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Key Concepts

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

Autonomous Differential Equations

An autonomous differential equation is one where the derivative y' depends only on the variable y, not explicitly on the independent variable t. This means the rate of change is determined solely by the current state y, making the direction field time-invariant and simplifying analysis of equilibrium points.

Equilibrium Solutions

Equilibrium solutions occur when the derivative y' equals zero, meaning the function y(t) remains constant over time. For autonomous equations y' = f(y), equilibrium solutions satisfy f(y0) = 0. These solutions correspond to horizontal lines in the direction field where the system is at rest.

Finding Equilibrium Points by Solving f(y) = 0

To find equilibrium solutions, set the right-hand side of the differential equation y' = f(y) equal to zero and solve for y. For example, in y' = 6 - 2y, setting 6 - 2y = 0 yields y = 3, which is the equilibrium solution where the system does not change.

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Related Practice

Textbook Question

Growth rate functions

a. Show that the logistic growth rate function f(P)=rP(1−P/K) has a maximum value of rK/4 at the point P=K/2.

Textbook Question

A second-order equation Consider the differential equation y''(t) - k²y(t) = 0 where k > 0 is a real number.

a. Verify by substitution that when k = 1, a solution of the equation is y(t) = C₁eᵗ + C₂e⁻ᵗ. You may assume this function is the general solution.

Textbook Question

42–43. Implicit solutions for separable equations For the following separable equations, carry out the indicated analysis.

a. Find the general solution of the equation.

y'(t) = t²/(y² + 1); y(−1) = 1, y(0) = 0, y(−1) = −1

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Textbook Question

27–30. Predator-prey models Consider the following pairs of differential equations that model a predator-prey system with populations x and y. In each case, carry out the following steps.

a. Identify which equation corresponds to the predator and which corresponds to the prey.

x′(t) = −3x + xy, y′(t) = 2y − xy

Textbook Question

23–26. Stirred tank reactions For each of the following stirred tank reactions, carry out the following analysis.

a. Write an initial value problem for the mass of the substance.

A 500-L tank is initially filled with pure water. A copper sulfate solution with a concentration of 20 g/L flows into the tank at a rate of 4 L/min. The thoroughly mixed solution is drained from the tank at a rate of 4 L/min.

Textbook Question

Solving Bernoulli equations Use the method outlined in Exercise 43 to solve the following Bernoulli equations.

a. y′(t) + y = 2y²

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