/*! 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} Q52 E Find all the solution of the sys... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all the solution of the system dx→dt=[λ10λ]x→ where is arbitrary constant.

Short Answer

Expert verified

The solution is x→(t)=c1eλt10+c2eλtt10+11.

Step by step solution

01

Explanation of the solution

Consider the system as follows.

dx→dt=λ10λx→dx→dt=Ax→

To find the eigenvalues of A as follows.

detA-μl=0λ-μ10λ-μ=0λ-μλ-μ-0=0λ-μ2=0

Simplify further as follows.

λ-μ=0λ=μ

Therefore, the eigenvalues are μ1=λand μ2=λ.

Now, to find the corresponding eigenvector as follows.

Av→=λv→λ10λv1v2=λv1v2λ±¹1+v2λ±¹2=λ±¹1λ±¹2v2=0,v1=arbitrary

Simplify further as follows.

v→=10

Since, the system has only on eigenvector, and find generalized vector which is a solution of a system (A-λ±ô)u→=v→as follows.

x→(t)=c1eλt10+c2eλtt10+11

02

Phase portrait of the solution

The phase portrait for λ<0is given in figure 1 as follows.

The phase portrait for λ<0is given in figure 2 as follows.

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 91Ó°ÊÓ!

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

See all solutions

Recommended explanations on Math Textbooks

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.