Explanation Given: The given system is X ′ = ( 0 1 1 0 ) X + ( 1 1 ) . Definition used: 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 = ( 0 1 1 0 ) and F ( t ) = ( 1 1 ) . Compute A 2 by multiplying matrix A with matrix A . A 2 = A × A = ( 0 1 1 0 ) ⋅ ( 0 1 1 0 ) = ( 1 0 0 1 ) = I Multiply matrix A 2 with matrix A and obtain A 3 . A 3 = A 2 × A = I × A = A Multiply the matrix A 3 with the matrix A to obtain the value of the matrix A 4 . A 4 = A 3 × A = A × A = A 2 = I Proceed in this way, the general value of a matrix raised to the power of a natural number is obtained as, A k = { A k = 1 , 3 , 5 , … I k = 2 , 4 , 6 , … 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 = I + A 1 ! t + I 2 ! t 2 + A 3 ! t 3 + … = I ( 1 + 1 2 ! t 2 + 1 4 ! t 4 + … ) + A ( t + 1 3 ! t 3 + 1 5 ! t 5 + … ) The expansion of the hyperbolic functions is as, cosh t = 1 + 1 2 ! t 2 + 1 4 ! t 4 + … sinh t = t + 1 3 ! t 3 + 1 5 ! t 5 + … Use these values of the expansion, it gives the value of e A t as, e A t = I ( cos ht ) + A ( sinh t ) = ( 1 0 0 1 ) ( cos ht ) + ( 0 1 1 0 ) ( sinh t ) = ( cos ht 0 0 cos ht ) + ( 0 sinh t sinh t 0 ) = ( cos ht sinh t sinh t cos ht ) Substitute – t for t and obtain e − A t . e − A t = ( cos h ( − t ) sinh ( − t ) sinh ( − t ) cos h ( − t ) ) = ( cos h ( t ) − sinh ( t ) − sinh ( t ) cos h ( t ) ) [ cos h ( − t ) = cos h ( t ) sinh ( − t ) = − sinh ( t ) ] The general solution is given by, X ( t ) = e A t C + e A t ∫ t 0 t e − A s F ( s ) d s

11th Edition

ISBN: 9781337652476

Author: ZILL

Publisher: CENGAGE L

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Chapter 8.4, Problem 11E

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