Explanation Given: The second order differential equation is θ ″ = ( cos θ − 0.5 ) sin θ , | θ | < π . Concept used: Saddle point: The condition for a critical point to be a saddle point is that τ 2 − 4 Δ > 0 and Δ < 0 where, τ is the trace of the matrix and Δ is the determinant of the matrix. Calculation: The second order differential equation is given as follows: θ ″ = ( cos θ − 0.5 ) sin θ … … ( 1 ) Substitute y for θ ′ in equation ( 1 ) . y ′ = ( cos θ − 0.5 ) sin θ The system of differential equation formed is as follows: y ′ = ( cos θ − 0.5 ) sin θ … … ( 2 ) θ ′ = y … … ( 3 ) Consider that θ ′ = P ( θ , y ) and y ′ = Q ( θ , y ) . Equation ( 2 ) and equation ( 3 ) are written as follows: Q ( θ , y ) = ( cos θ − 0.5 ) sin θ P ( θ , y ) = y The critical point ( θ , y ) is the point for which θ ′ = 0 and y ′ = 0 . Solve the equation θ ′ = 0 as follows: θ ′ = 0 y = 0 Solve the equation y ′ = 0 as follows: y ′ = 0 ( cos θ − 0.5 ) sin θ = 0 From the above equation it is observed that either the value of sin θ is 0 or the value of ( cos θ − 0.5 ) is 0 . Solve the equation sin θ = 0 as follows: sin θ = 0 θ = 0 ( | θ | < π ) Solve the equation cos θ − 0.5 = 0 as follows: cos θ − 0.5 = 0 cos θ = 0.5 θ = ± π 3 ( | θ | < π ) Therefore, the critical points are ( 0 , 0 ) , ( 0 , − π 3 ) and ( 0 , π 3 ) . Differentiate P ( θ , y ) partially with respect to θ as follows: ∂ P ( θ , y ) ∂ θ = ∂ ( y ) ∂ θ = 0 Differentiate P ( θ , y ) partially with respect to y as follows: ∂ P ( θ , y ) ∂ y = ∂ ( y ) ∂ y = 1 Differentiate Q ( θ , y ) partially with respect to θ as follows: ∂ Q ( θ , y ) ∂ θ = ∂ ( ( cos θ − 0.5 ) sin θ ) ∂ θ = ∂ ( cos θ sin θ − 0.5 sin θ ) ∂ θ = ( cos θ ) 2 − ( sin θ ) 2 − 0.5 cos θ = cos ( 2 θ ) − 0.5 cos θ Differentiate Q ( θ , y ) partially with respect to y as follows: ∂ Q ( θ , y ) ∂ y = ∂ ( ( cos θ − 0.5 ) sin θ ) ∂ y = 0 The Jacobian matrix formed for the given system is as follows: g ′ ( θ , y ) = ( ∂ P ∂ θ ∂ P ∂ y ∂ Q ∂ θ ∂ Q ∂ y ) = ( 0 1 cos ( 2 θ ) − cos θ 2 0 ) For the critical point ( 0 , 0 ) the Jacobian matrix is as follows: g ′ ( 0 , 0 ) = ( 0 1 1 2 0 ) Consider the trace of the above matrix as τ

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 11.3, Problem 21E

To determine

The nature of the critical points for the second order differential equation θ″=(cosθ−0.5)sinθ, |θ|<π.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 11.3, Problem 20E

Chapter 11.3, Problem 22E

Students have asked these similar questions

In Problems 1–4, state the order of the given ordinary differential equation. Also determine if the equation is linear or not

For each dif erential equation in Problems 1–21, find the general solutionby finding the homogeneous solution and a particular solution. Please DO NOT YOU THE PI method where 1/f(r) * x. Dont do that. Instead do this, assume for yp = to something, do the 1 and 2 derivative of it and then plug it in the equation to find the answer.

III. Solutions of Differential Equations. Verify the following

Chapter 11 Solutions

Advanced Engineering Mathematics

Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - Prob. 15E Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.2 - Prob. 17E Ch. 11.2 - Prob. 18E Ch. 11.2 - Prob. 19E Ch. 11.2 - Prob. 20E Ch. 11.2 - Prob. 21E Ch. 11.2 - Prob. 22E Ch. 11.2 - Prob. 23E Ch. 11.2 - Prob. 24E Ch. 11.2 - Prob. 25E Ch. 11.2 - Prob. 26E Ch. 11.3 - Prob. 1E Ch. 11.3 - Prob. 2E Ch. 11.3 - Prob. 3E Ch. 11.3 - Prob. 4E Ch. 11.3 - Prob. 5E Ch. 11.3 - Prob. 6E Ch. 11.3 - Prob. 7E Ch. 11.3 - Prob. 8E Ch. 11.3 - Prob. 9E Ch. 11.3 - Prob. 10E Ch. 11.3 - Prob. 11E Ch. 11.3 - Prob. 12E Ch. 11.3 - Prob. 13E Ch. 11.3 - Prob. 14E Ch. 11.3 - Prob. 15E Ch. 11.3 - Prob. 16E Ch. 11.3 - Prob. 17E Ch. 11.3 - Prob. 18E Ch. 11.3 - Prob. 19E Ch. 11.3 - Prob. 20E Ch. 11.3 - Prob. 21E Ch. 11.3 - Prob. 22E Ch. 11.3 - Prob. 23E Ch. 11.3 - Prob. 24E Ch. 11.3 - Prob. 25E Ch. 11.3 - Prob. 26E Ch. 11.3 - Prob. 27E Ch. 11.3 - Prob. 28E Ch. 11.3 - Prob. 29E Ch. 11.3 - Prob. 30E Ch. 11.3 - Prob. 31E Ch. 11.3 - Prob. 32E Ch. 11.3 - Prob. 33E Ch. 11.3 - Prob. 34E Ch. 11.3 - Prob. 35E Ch. 11.3 - Prob. 36E Ch. 11.3 - Prob. 37E Ch. 11.3 - Prob. 38E Ch. 11.3 - Prob. 39E Ch. 11.4 - Prob. 1E Ch. 11.4 - Prob. 2E Ch. 11.4 - Prob. 3E Ch. 11.4 - Prob. 4E Ch. 11.4 - Prob. 5E Ch. 11.4 - Prob. 6E Ch. 11.4 - Prob. 7E Ch. 11.4 - Prob. 8E Ch. 11.4 - Prob. 9E Ch. 11.4 - Prob. 11E Ch. 11.4 - Prob. 12E Ch. 11.4 - Prob. 13E Ch. 11.4 - Prob. 14E Ch. 11.4 - Prob. 15E Ch. 11.4 - Prob. 16E Ch. 11.4 - Prob. 17E Ch. 11.4 - Prob. 18E Ch. 11.4 - Prob. 19E Ch. 11.4 - Prob. 20E Ch. 11.4 - Prob. 21E Ch. 11.4 - Prob. 22E Ch. 11 - Prob. 1CR Ch. 11 - Prob. 2CR Ch. 11 - Prob. 3CR Ch. 11 - Prob. 4CR Ch. 11 - Prob. 5CR Ch. 11 - Prob. 6CR Ch. 11 - Prob. 7CR Ch. 11 - Prob. 8CR Ch. 11 - Prob. 11CR Ch. 11 - Prob. 12CR Ch. 11 - Prob. 13CR Ch. 11 - Prob. 15CR Ch. 11 - Prob. 16CR Ch. 11 - Prob. 17CR

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

The nature of the critical points for the second order differential equation θ ″ = ( cos θ − 0.5 ) sin θ , | θ | < π .

Solution Summary: The author explains the critical points for the second order differential equation theta′′=(mathrmcos

Concept explainers

Videos

The nature of the critical points for the second order differential equation θ″=(cosθ−0.5)sinθ, |θ|<π.

Want to see the full answer?

Chapter 11 Solutions

The nature of the critical points for the second order differential equation θ ″ = ( cos θ − 0.5 ) sin θ , | θ | &lt; π .

Solution Summary: The author explains the critical points for the second order differential equation theta′′=(mathrmcos

Concept explainers

Videos

The nature of the critical points for the second order differential equation θ″=(cosθ−0.5)sinθ, |θ|<π.

Want to see the full answer?

Chapter 11 Solutions

The nature of the critical points for the second order differential equation θ ″ = ( cos θ − 0.5 ) sin θ , | θ | < π .