Explanation Given: The given system is, d x d t = 3 x + 2 y + 4 z d y d t = 2 x + 2 z d z d t = 4 x + 2 y + 3 z Calculation: For the given system, the coefficient matrix is A = ( 3 2 4 2 0 2 4 2 3 ) . The characteristic equation for the matrix A is | ( A − λ I ) | = 0 where I is the identity matrix. Substitute the values of matrix A and I in above expression. | 3 − λ 2 4 2 − λ 2 4 2 3 − λ | = 0 ( 3 − λ ) [ ( λ − 3 ) λ − 4 ] − 2 [ 6 − 2 λ − 8 ] + 4 [ 4 + 4 λ ] = 0 ( 3 − λ ) [ λ 2 − 3 λ − 4 ] − 2 [ − 2 − 2 λ ] + 16 ( 1 + λ ) = 0 ( 3 − λ ) ( λ − 4 ) ( λ + 1 ) + 4 ( 1 + λ ) + 16 ( 1 + λ ) = 0 Simplify further, ( λ + 1 ) [ ( 3 − λ ) ( λ − 4 ) + 4 + 16 ] = 0 ( λ + 1 ) [ − 12 + 7 λ − λ 2 + 20 ] = 0 ( λ + 1 ) ( 8 + 7 λ − λ 2 ) = 0 ( λ + 1 ) ( 8 − λ ) ( λ + 1 ) = 0 Therefore, the eigenvalues of the above system is λ 1 = 8 and λ 2 = − 1 with multiplicity 2 Obtain the eigenvector K 1 corresponding to the eigenvalue λ 1 = 8 . Solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 z 1 ) . Substitute the values of A , λ 1 and K 1 in the above equation gives, [ ( 3 2 4 2 0 2 4 2 3 ) − ( 8 0 0 0 8 0 0 0 8 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( − 5 2 4 2 − 8 2 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Now replace the second Row with the first Row, ( 2 − 8 2 − 5 2 4 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) By applying the row transformation R 1 → 1 2 R 1 in the above system of equations, ( 1 − 4 1 − 5 2 4 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again by row transformation, R 2 → R 2 + 5 R 1 ( 1 − 4 1 0 − 18 9 4 2 − 5 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again row transformation, R 3 → R 3 − 4 R 1 , ( 1 − 4 1 0 − 18 9 0 18 − 9 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Again by row transformation, R 3 → R 3 + R 2 , ( 1 − 4 1 0 − 18 9 0 0 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( x 1 − 4 y 1 + z 1 − 18 y 1 + 9 z 1 0 ) = ( 0 0 0 ) Which is equivalent to the equations, x 1 − 4 y 1 + z 1 = 0 and − 18 y 1 + 9 z 1 = 0 . On solving further it gives, x 1 = 4 y 1 − z 1 and y 1 = 1 2 z 1 which implies, x 1 = 4 2 z 1 − z 1 x 1 = 2 z 1 − z 1 x 1 = z 1 Assume that z 1 = p 1 for some scalar p 1 which implies, x 1 = p 1 y 1 = 1 2 p 1 Therefore, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( p 1 1 2 p 1 p 1 ) = p 1 ( 1 1 2 1 ) = 2 p 1 ( 2 1 2 ) Thus, the eigenvector K 1 corresponding to λ 1 = 8 is ( 2 1 2 )

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

See similar textbooks

Videos

Question

Chapter 8.2, Problem 26E

To determine

To find: The general solution of the given system.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.2, Problem 25E

Chapter 8.2, Problem 27E

Students have asked these similar questions

In Problems 39–43, solve each system of equations. Į 2r + y + 3 = 0 x² + y? = 5 x + y? = 6y x = 3y S2xy + y? = 10 39. 40. 41. | 3y² – xy = 2 Ј Зx? + 4ху + 5у? 3D 8 42. x² - 3x + y² + y = -2 43. lx² + 3xy + 2y² = 0 + y + 1 = 0 y

1. Y = 5x + 2 2. 5y = 25x + 10 %3D 3. 5y = 10x + 10 4. 2y = 10x – 4 Which pair of equations forms a system that has infinitely many solutions? 2 and 4 2 and 3 O 1 and 2 O 1 and 4

In Problems 19–56, solve each system of equations. If the system has no solution, say that it is inconsistent. For Problems 19–30, graphthe lines of the system

Chapter 8 Solutions

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Knowledge Booster

Learn more about

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

The general solution of the given system.

Solution Summary: The author explains that the general solution of the given system is X(t)=c_1 left, where I is the identity matrix. Substitute the values of

Videos

To find: The general solution of the given system.

Want to see the full answer?

Chapter 8 Solutions