Explanation Given: The given first-order linear differential equations are, y 1 ′ = 6 y 1 − y 2 + 2 y 3 y 2 ′ = 3 y 2 − 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 in the variables y = P w and y ′ = P w ′ . Calculation: Consider the given first-order linear differential equations: y 1 ′ = 6 y 1 − y 2 + 2 y 3 y 2 ′ = 3 y 2 − y 3 y 3 ′ = y 3 Rewrite the equations in the matrix form, we get [ y 1 ′ y 2 ′ y 3 ′ ] = [ 6 − 1 2 0 3 − 1 0 0 1 ] [ y 1 y 2 y 3 ] Find the eigenvalues and eigenvectors. The characteristics equation is, det ( [ 6 − 1 2 0 3 − 1 0 0 1 ] − λ [ 1 0 0 0 1 0 0 0 1 ] ) = 0 det ( [ 6 − 1 2 0 3 − 1 0 0 1 ] − [ λ 0 0 0 λ 0 0 0 λ ] ) = 0 | 6 − λ − 1 2 0 3 − λ − 1 0 0 1 − λ | = 0 ( 6 − λ ) | 3 − λ − 1 0 1 − λ | = 0 Further simplify as: ( 6 − λ ) ( 3 − λ ) ( 1 − λ ) = 0 λ = 6 , 3 , 1 The eigenvalues are λ 1 = 6 , λ 2 = 3 and λ 3 = 1 . Find the eigenvector corresponding to eigenvalue λ = 6 by solving the homogeneous system ( A − 6 I ) X = 0 : Let X = [ x 1 x 2 x 3 ] . ( [ 6 − 1 2 0 3 − 1 0 0 1 ] − 6 [ 1 0 0 0 1 0 0 0 1 ] ) [ x 1 x 2 x 3 ] = 0 ( [ 6 − 1 2 0 3 − 1 0 0 1 ] − [ 6 0 0 0 6 0 0 0 6 ] ) [ x 1 x 2 x 3 ] = 0 [ 0 − 1 2 0 − 3 − 1 0 0 − 5 ] [ x 1 x 2 x 3 ] = 0 It reduces as: [ 0 − 1 2 0 − 3 − 1 0 0 − 5 ] → [ 0 1 0 0 0 0 0 0 1 ] 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 ( A − 3 I ) X = 0 : Let X = [ x 1 x 2 x 3 ] .

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ISBN: 9780357156100

Author: Ron Larson

Publisher: Cengage Limited

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Chapter 7.CR, Problem 80CR

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To find:

The solutions for the first-order linear differential equations

y1′=6y1−y2+2y3y2′= 3y2−y3y3′= y3.

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Chapter 7.CR, Problem 79CR

Chapter 7.CR, Problem 81CR

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Chapter 7 Solutions

Elementary Linear Algebra

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The solutions for the first-order linear differential equations y 1 ′ = 6 y 1 − y 2 + 2 y 3 y 2 ′ = 3 y 2 − y 3 y 3 ′ = y 3 .

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

Videos

To find:

The solutions for the first-order linear differential equations

y1′=6y1−y2+2y3y2′= 3y2−y3y3′= y3.

Want to see the full answer?

Chapter 7 Solutions

The solutions for the first-order linear differential equations y 1 ′ = 6 y 1 − y 2 + 2 y 3 y 2 ′ = 3 y 2 − y 3 y 3 ′ = y 3 .

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

Videos

To find: The solutions for the first-order linear differential equations y1′=6y1−y2+2y3y2′= 3y2−y3y3′= y3.

Want to see the full answer?

Chapter 7 Solutions

To find:

The solutions for the first-order linear differential equations

y1′=6y1−y2+2y3y2′= 3y2−y3y3′= y3.