Explanation Given: The given first-order linear differential equations are: y 1 ′ = − 2 y 1 + y 3 y 2 ′ = 3 y 2 + 4 y 3 y 3 ′ = y 3 Procedure used: A system of first-order linear differential equations has the form y 1 ′ = a 11 y 1 + a 12 y 2 + ⋯ + a 1 n y n y 2 ′ = a 21 y 1 + a 22 y 2 + ⋯ + a 2 n y n ⋮ y n ′ = a n 1 y 1 + a n 2 y 2 + ⋯ + a n n y n Where each y i is a function of t and y i ′ = d y i d t . If you let y ′ = [ y 1 ′ y 2 ′ ⋮ y n ′ ] , y = [ y 1 y 2 ⋮ y n ] and A = [ a 11 a 12 ⋯ a n 1 a 21 a 22 ⋯ a n 2 ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ] Then the system can be written in matrix form as y ′ = A y Then find the matrix P that diagonalizes A . Apply the change the variables y = P w and y ′ = P w ′ . Calculation: Consider the given first-order linear differential equations: y 1 ′ = − 2 y 1 + y 3 y 2 ′ = 3 y 2 + 4 y 3 y 3 ′ = y 3 . Rewrite the equations in the matrix form, we get [ y 1 ′ y 2 ′ y 3 ′ ] = [ − 2 0 1 0 3 4 0 0 1 ] [ y 1 y 2 y 3 ] Find the eigenvalues and eigenvectors: The characteristics equation is: | λ [ 1 0 0 0 1 0 0 0 1 ] − [ − 2 0 1 0 3 4 0 0 1 ] | = 0 | [ λ 0 0 0 λ 0 0 0 λ ] − [ − 2 0 1 0 3 4 0 0 1 ] | = 0 | [ λ + 2 0 − 1 0 λ − 3 − 4 0 0 λ − 1 ] | = 0 ( λ + 2 ) ⋅ | λ − 3 − 4 0 λ − 1 | = 0 Further simplify as: ( λ + 2 ) ( λ − 3 ) ( λ − 1 ) = 0 λ = − 2 , 3 , 1 The eigenvalues are λ 1 = − 2 , λ 2 = 3 and λ 3 = 1 . Find the eigenvector corresponding to eigenvalue λ = − 2 by solving the homogeneous system ( − 2 I − A ) X = 0 : Let X = [ x 1 x 2 x 3 ] . ( − 2 [ 1 0 0 0 1 0 0 0 1 ] − [ − 2 0 1 0 3 4 0 0 1 ] ) [ x 1 x 2 x 3 ] = 0 ( [ − 2 0 0 0 − 2 0 0 0 − 2 ] − [ − 2 0 1 0 3 4 0 0 1 ] ) [ x 1 x 2 x 3 ] = 0 [ 0 0 − 1 0 − 5 − 4 0 0 − 3 ] [ x 1 x 2 x 3 ] = 0 It indicate as, x 2 = 0 , x 3 = 0 and x 1 is free. Choose the arbitrary value x 1 = 1 . The eigenvector is: p 1 = [ 1 0 0 ] Find the eigenvector corresponding to eigenvalue λ = 3 by solving the homogeneous system ( 3 I − A ) X = 0 : Let X = [ x 1 x 2 x 3 ]

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN: 9781305658004

Author: Ron Larson

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 7.4, Problem 28E

To determine

To find:

The solutions for the given first-order linear differential equations y1′=−2y1 +y3y2′= 3y2+4y3y3′= y3.

Expert Solution & Answer

Chapter 7 Solutions

Elementary Linear Algebra (MindTap Course List)

Ch. 7.1 - Verifying Eigenvalues and Eigenvectors in...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and Eigenvectors in...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Verifying Eigenvalues and EigenvectorsIn Exercises...Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Determining Eigenvectors In Exercise 9-12,...Ch. 7.1 - Determining Eigenvectors In Exercise 9-12,...

Knowledge Booster

$Algebra$

Learn more about

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

The solutions for the given first-order linear differential equations y 1 ′ = − 2 y 1 + y 3 y 2 ′ = 3 y 2 + 4 y 3 y 3 ′ = y 3 .

Solution Summary: The author explains the solution of the given first-order linear differential equations.

Videos

To find:

The solutions for the given first-order linear differential equations y1′=−2y1 +y3y2′= 3y2+4y3y3′= y3.

Chapter 7 Solutions

The solutions for the given first-order linear differential equations y 1 ′ = − 2 y 1 + y 3 y 2 ′ = 3 y 2 + 4 y 3 y 3 ′ = y 3 .

Solution Summary: The author explains the solution of the given first-order linear differential equations.

Videos

To find: The solutions for the given first-order linear differential equations y1′=−2y1 +y3y2′= 3y2+4y3y3′= y3.

Chapter 7 Solutions

To find:

The solutions for the given first-order linear differential equations y1′=−2y1 +y3y2′= 3y2+4y3y3′= y3.