(a) Explanation Consider the following differential equation: y " + 2 y ' + 2 y = cos t + δ ( t − π 2 ) With the initial conditions are y ( 0 ) = 0 , y ' ( 0 ) = 0 . The objective is to find the solution of the above initial value problem. Take Laplace transform on both sides of the equation y " + 2 y ' + 2 y = cos t + δ ( t − π 2 ) . L { y " + 2 y ' + 2 y } = L { cos t + δ ( t − π 2 ) } L { y " } + L { 2 y ' } + L { 2 y } = L { cos t } + L { δ ( t − π 2 ) } To simplify L { y " } + L { 2 y ' } + L { 2 y } = L { cos t } + L { δ ( t − π 2 ) } using the linearity property. ( s 2 Y ( s ) − s y ( 0 ) − y ' ( 0 ) ) + [ 2 s Y ( s ) − 2 y ( 0 ) ] + 2 Y ( s ) = s s 2 + 1 + e − ( π 2 s ) s 2 Y ( s ) − s y ( 0 ) − y ' ( 0 ) + 2 s Y ( s ) − 2 y ( 0 ) + 2 Y ( s ) = s s 2 + 1 + e − ( π 2 s ) Substitute the value of y ( 0 ) = 0 , y ' ( 0 ) = 0 in s 2 Y ( s ) − s y ( 0 ) − y ' ( 0 ) + 2 s Y ( s ) − 2 y ( 0 ) + 2 Y ( s ) = s s 2 + 1 + e − ( π 2 s ) . s 2 Y ( s ) − s × 0 − 0 + 2 s Y ( s ) − 0 + 2 Y ( s ) = s s 2 + 1 + e − ( π 2 s ) ( s 2 + 2 s + 2 ) Y ( s ) = s s 2 + 1 + e − ( π 2 s ) Y ( s ) = s ( s 2 + 1 ) ( s 2 + 2 s + 2 ) + e − ( π 2 s ) ( s 2 + 2 s + 2 ) The partial fraction decomposition of s ( s 2 + 1 ) ( s 2 + 2 s + 2 ) is as follows: s ( s 2 + 1 ) ( s 2 + 2 s + 2 ) = ( 1 5 ) s + ( 2 5 ) ( s 2 + 1 ) + ( − 1 5 ) s + ( − 4 5 ) ( s 2 + 2 s + 2 ) = s + 2 5 ( s 2 + 1 ) − s + 4 5 ( s 2 + 2 s + 2 ) = s + 2 5 ( s 2 + 1 ) − s + 3 + 1 5 ( s 2 + 2 s + 2 ) = 1 5 s ( s 2 + 1 ) + 2 5 1 ( s 2 + 1 ) − 1 5 (a) To determine To draw: The graph of the solution.

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 6.5, Problem 11P

(a)

To determine

The solution of the given initial value problem.

(a)

To determine

To draw: The graph of the solution.

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Chapter 6 Solutions

Elementary Differential Equations

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