/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the equilibrium solution of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 19

Find the equilibrium solution of the following equations, make a sketch of the direction field, for \(t \geq 0,\) and determine whether the equilibrium solution is stable. The direction field needs to indicate only whether solutions are increasing or decreasing on either side of the equilibrium solution. $$y^{\prime}(t)=-\frac{y}{3}-1$$

Short Answer

Expert verified
Answer: The equilibrium solution for the given first-order differential equation is \(y = -3\), and it is stable.

Step by step solution

01

Find the equilibrium solution

To find the equilibrium solution, we need to set the derivative equal to zero: $$y^{\prime}(t) = 0$$ Substitute the given equation: $$0 = -\frac{y}{3} - 1$$ Solve for \(y\): $$y = -3$$ The equilibrium solution is \(y = -3\).
02

Determine the stability of the equilibrium solution

To determine the stability, we will analyze the sign of the derivative. Analyzing the given equation, we can notice that when \(y > -3\), the derivative is negative, and when \(y < -3\), the derivative is positive. This implies that the solutions will approach the equilibrium solution as time goes on. Therefore, the equilibrium solution \(y = -3\) is stable.
03

Sketch the direction field

To sketch the direction field, we will evaluate some points around the equilibrium solution to find out whether the solutions are increasing or decreasing on either side of the equilibrium solution. - When \(y > -3\), the derivative is negative, meaning the solutions are decreasing. Therefore, the arrow will point down on the right side of the equilibrium solution in the direction field. - When \(y < -3\), the derivative is positive, meaning the solutions are increasing. Therefore, the arrow will point up on the left side of the equilibrium solution in the direction field. In summary, the equilibrium solution is \(y = -3\) and is stable, with solutions decreasing when \(y > -3\) and increasing when \(y < -3\).

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 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of \(t.\) Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t.\) Use this idea to solve the following initial value problems. $$e^{-t} y^{\prime}(t)-e^{-t} y=e^{2 t}, y(0)=4$$

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. A pot of boiling soup \(\left(100^{\circ} \mathrm{C}\right)\) is put in a cellar with a temperature of \(10^{\circ} \mathrm{C}\). After 30 minutes, the soup has cooled to \(80^{\circ} \mathrm{C}\). When will the temperature of the soup reach \(30^{\circ} \mathrm{C} ?\)

Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.

Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.