Explanation Given: The given system is X ′ = ( 1 0 0 2 ) X + ( t e 4 t ) . Concept: For any matrix A of order n × n , the expansion of e A t in power of the matrix A is, e A t = 1 + A t+ A 2 2 ! t 2 + A 3 3 ! t 3 + … + A k k ! t k + … . The general solution of the system X ′ = A X + F ( t ) is given by X ( t ) = e A t C + e A t ∫ t 0 t e − A s F ( s ) d s where C is constant. Calculation: In the given system, the matrix is A = ( 1 0 0 2 ) and F ( t ) = ( t e 4 t ) . Compute A 2 by multiplying matrix A with matrix A . A 2 = A × A = ( 1 0 0 2 ) ⋅ ( 1 0 0 2 ) = ( 1 0 0 4 ) Find the matrix A 3 by multiplying matrix A 2 with matrix A . A 3 = A 2 × A = ( 1 0 0 8 ) Thereafter, to obtain the value of the matrix A 4 , multiply the matrix A 3 with the matrix A,. A 4 = A 3 × A = ( 1 0 0 16 ) From the above calculation, the matrices can be written as follows. A = ( 1 0 0 2 1 ) , A 2 = ( 1 0 0 2 2 ) , A 3 = ( 1 0 0 2 3 ) , A 4 = ( 1 0 0 2 4 ) , … Proceed in this way, the general value of a matrix raised to the power of a natural number is obtained as, A k = ( 1 0 0 2 k ) for k = 1 , 2 , 3 , … Thus, the value of the expansion e A t in power of the matrix A is e A t = 1 + A 1 ! t + A 2 2 ! t 2 + A 3 3 ! t 3 + … Substitute the value of matrices in above expression, e A t = ( 1 0 0 1 ) + 1 1 ! ( 1 0 0 2 ) t + 1 2 ! ( 1 0 0 4 ) t 2 + 1 3 ! ( 1 0 0 8 ) t 3 + … = ( 1 + t + t 2 2 ! + t 3 3 ! + … 0 0 1 + 2 t + ( 2 t ) 2 2 ! + ( 2 t ) 3 3 ! + … ) = ( e t 0 0 e 2 t ) Substitute – t for t in the expansion of e A t and obtain e − A t

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

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.4, Problem 10E

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To find: The general solution of the given system using matrix A.

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Chapter 8.4, Problem 9E

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

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