Log In Start studying!

Select your language

Suggested languages for you:

Americas

English (US)

Europe

Q5.4-29E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 272

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later

Short Answer

Find the critical points and solve the related phase plane differential equation for the system $\frac{dx}{dt} = (x - 1) (y - 1), \frac{dy}{dt} = y (y - 1)$ .Describe (without using computer software) the asymptotic behavior of trajectories (as role="math" localid="1664302603010" $t \to \infty$ ) that start at
(3, 2)
(2, 1/2)
( -2, 1/2)
(3, -2)

a. The solution approaches infinity.

b. The solution approaches point (1,0).

c. The solution approaches point (1,0).

d. The solution approaches point (1,0).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Find the critical point

Here the equation is:

$\frac{d x}{d t} = (x - 1) (y - 1) \frac{d y}{d t} = y (y - 1)$

If y=0 then x=1.

And in (2) ( $y \neq 0$ ) the y=1,

This is satisfied for all numbers x. So, all points are of the form (x, 1). So, the critical point is (1, 0),(x,1).

Find the value of y

Here,

$\frac{dy}{dx} = \frac{y}{x - 1}$

Solve this by variable separating.

Y=C(x-1).

Now implied that $x - 1 \neq 0$ so that could have it in the denominator. The solution x=1 makes the system.

$\frac{d x}{d t} = 0 \frac{d y}{d x} = y (y - 1)$

This curve x=1 is also a solution of the system because when we differentiate it by t so, $\frac{d x}{d t} = 0$ .so the value of y is y=C(x-1) or x=1.

Solve for points (3,2)

The solution that starts at (3, 2) is above the lines of critical points (x, 1) and it is also right of the x=1. that means that the corresponding trajectory is of the form y=C(x-1). So put x=3, y=2 then C=1.

So, the trajectory line is y=x-1.

So, the system is for y=1.

$\frac{dx}{dt} > 0 \frac{dy}{dx} > 0$

So, the solution along the trajectory starting at (3,2) is increasing and therefore going to infinity as $t \to \infty$ .

Evaluate for point (2,12)

Put x=1 and y=1/2 then C=1/2.

The trajectory of this solution is $y = \frac{x - 1}{2}$ .

Since one is below the line y=1 line and above y=0 line that y-1<0 and y>0. So,

$\frac{dx}{dt} < 0 \frac{dy}{dx} < 0$

That means the trajectory decreases until it hits the critical point (1,0).

Solve for point (-2,12).

Put x=-2, y=1/2 then C=-1/6.

The trajectory of this solution is $y = - \frac{x - 1}{6}$ .

Since one is below the line y=1 line and above y=0 line that y-1<0 and y>0. So,

$\frac{dx}{dt} > 0 \frac{dy}{dx} < 0$

That means the trajectory increases along the x-axis and decreases along the y-axis until it hits the critical point (1,0).

Determine the value for point (3,-2).

Put x = 3, y = -2 then C = -1.

The trajectory of this solution is $y = - (x - 1)$ .

Thus, one is below the line y=1 line and above y = 0 line that y -1 < 0 and y > 0. So,

$\frac{dx}{dt} < 0 \frac{dy}{dx} > 0$

That means the trajectory decreases along the x-axis and increases along the y-axis until it hits the critical point (1,0).

Most popular questions for Math Textbooks

Using the software, sketch the direction field in the phase-plane for the system $\frac{d x}{d t} = - 2 x + y, \frac{d y}{d t} = - 5 x - 4 y$ . From the sketch, predict the asymptotic limit (as $t \to \infty$

of the solution starting at (1, 1).

Find all the critical points of the system

$\frac{d x}{d t} = x^{2} - 1 \frac{d y}{d t} = x y$

and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

In Problem 31, 3 L/min of liquid flowed from tank A into tank B and 1 L/min from B to A. Determine the mass of salt in each tank at time $t ⩾ 0$ if, instead, 5 L/min flows from A into B and 3 L/min flows from B into A, with all other data the same.

In Problems 10–13, use the vectorized Euler method with = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$y'' + ty' + y = 0; y (0) = 1, y' (0) = 0 on [0, 1]$

In Problems 19–24, convert the given second-order equation into a first-order system by setting v=y’. Then find all the critical points in the yv-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12).

$\frac{d^{2} y}{d t^{2}} + y = 0$

Want to see more solutions like these?

Sign up for free to discover our expert answers

Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free