Explanation The given non-homogeneous system is, X ′ = ( 3 2 − 2 − 1 ) X + ( 1 1 ) Calculation: The system is given in the form X ′ = A X + F ( t ) (1). Consider, the characteristic equation of a matrix A as | A − λ I | = 0 , where I is identity matrix. Now, the characteristic equation of the coefficient matrix A . is | ( A − λ I ) | = 0 | 3 − λ 2 − 2 − 1 − λ | = 0 ( 3 − λ ) ( − 1 − λ ) + 4 = 0 − 3 − 3 λ + λ + λ 2 + 4 = 0 Simplifying the above obtained expression gives, λ 2 − 2 λ + 1 = 0 ( λ − 1 ) 2 = 0 λ = 1 , 1 So, Eigen values are λ 1 = 1 , λ 2 = 1 , . Now evaluating the Eigen vector for λ 1 = 1 , ( A − I 0 ) = ( 2 2 0 − 2 − 2 0 ) On solving, it gives 2 K 1 + 2 K 2 = 0 Which on further evaluating gives K 1 = − K 2 . Thus, the corresponding Eigen vector is K = ( 1 − 1 ) . Hence, the value of corresponding solution vector is X 1 = ( 1 − 1 ) e t . Since, there are two repeated Eigen value, so consider another Eigen vector as P = ( a b ) . Now, solve the next system ( A − 2 I ) P = K to get the value of another Eigen vector. ( A − 2 I ) ( a b ) = K ( 2 2 − 2 − 2 ) ( a b ) = ( 1 − 1 ) Apply the operation R 1 → R 1 + R 2 in the above expression. ( 2 2 0 0 ) ( a b ) = ( 1 0 ) On solving, it gives 2 a + 2 b = 1 In the above obtained expression, there is only one equation and two variables, so consider a = 0 , then b becomes 1 2 . Thus, the corresponding Eigen vector is P = ( 0 1 2 ) . Hence, the value of corresponding solution vector is, X 2 = K t e λ 1 t + P e λ 2 t = ( 1 − 1 ) t e t + ( 0 1 2 ) e t . Consider the coefficient matrix A having real Eigen values of homogeneous system, K be an Eigen vector and X 1 , X 2 be the solution vectors of the corresponding Eigen values then, the value of complementary function X c is, X c = C 1 X 1 + C 2 X 2 Substituting the value of X 1 and X 2 in above expression it gives, X c = C 1 ( 1 − 1 ) t e t + C 2 [ ( 1 − 1 ) t e t + ( 0 1 2 ) e t ] (2) Now, the entries of X 1 form the first column of solution matrix ϕ ( t ) , and the entries of X 2 form the second column of ϕ ( t ) which means the value of ϕ ( t ) is ϕ ( t ) = ( e t t e t − e t 1 2 e t − t e t ) Evaluating determinant of the above matrix ϕ ( t ) , det [ ϕ ( t ) ] = | e t t e t − e t 1 2 e t − t e t | = e t ( 1 2 e t − t e t ) − t e t ( − e t ) = 1 2 e 2 t − t e 2 t + t e 2 t = 1 2 e 2 t Now, the value of the value of ϕ − 1 ( t ) is ϕ − 1 ( t ) = 1 | ϕ ( t ) | ⋅ ( Adj

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

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 8.3, Problem 22E

To determine

To solve: The given non-homogeneous system by using variation of parameters.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.3, Problem 21E

Chapter 8.3, Problem 23E

Students have asked these similar questions

#6 only

5.2. Solve the problem 0 0 Uzz = 0 u (x,0) = uz (r, 0) = 0 uz (0, t) = 1, u (1, t) 0 Utt %3! %3D t2 0.

3. 2хydx - (3xу + 2y?)dy %3D0 o (x - 2y)*(2x +y) = c (х — у)"(х + у) %3 с (х + 2y) (2х- у)* %3 с (x – 2y)* = c(2x + y)

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 given non-homogeneous system by using variation of parameters.

Solution Summary: The author explains how to solve the given non-homogeneous system by using variation of parameters. The characteristic equation of a matrix A is left|A-lambda Iright|=

Concept explainers

Videos

To solve: The given non-homogeneous system by using variation of parameters.

Want to see the full answer?

Chapter 8 Solutions