Explanation Given: The initial value is problem is, y ′ ′ + y = e − t , y ( 0 ) = 0 , y ′ ( 0 ) = 1 Approach: The Laplace of n th derivative of function is given by, L { f n ( t ) } = s n L { f ( t ) } − s n − 1 f ( 0 ) − ... s f ( n − 2 ) ( 0 ) − f ( n − 1 ) ( 0 ) ... ( 1 ) The Laplace of e − t function is given by, L { e − t } = 1 s + 1 ... ( 2 ) By the Convolution Theorem, L − 1 [ F ( s ) G ( s ) ] = L − 1 [ F ( s ) ] ∗ L − 1 [ G ( s ) ] By the definition of Convolution product theorem, ( f ∗ g ) ( t ) = ∫ 0 t f ( t − τ ) g ( τ ) d τ Calculation: Consider the given differential equation. y ′ ′ + y = e − t Transform the above differential equation on both sides by using the equations ( 1 ) and ( 2 ) . s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) + Y ( s ) = 1 s + 1 Apply the given initial conditions y ( 0 ) = 0 and y ′ ( 0 ) = 1 on above transformed equation. s 2 Y ( s ) − 0 − 1 + Y ( s ) = 1 s + 1 s 2 Y ( s ) + Y ( s ) − 1 = 1 s + 1 Solve the above equation for Y ( s ) . s 2 Y ( s ) + Y ( s ) = 1 s + 1 + 1 Y ( s ) ( s 2 + 1 ) = 1 s + 1 + 1 Y ( s ) = 1 ( s 2 + 1 ) ( s + 1 ) + 1 ( s 2 + 1 ) Find the inverse transform of the above equation. L − 1 [ Y ( s ) ] = L − 1 [ 1 ( s 2 + 1 ) ( s + 1 ) ] + L − 1 [ 1 ( s 2 + 1 ) ] = L − 1 [ 1 ( s 2 + 1 ) ( s + 1 ) ] + sin t … ( 3 ) From the above equation, take the term L − 1 [ 1 ( s 2 + 1 ) ( s + 1 ) ] and express as a convolution integral

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ISBN: 9780321990167

Author: ANNIN

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 10.9, Problem 26P

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Chapter 10 Solutions

EBK DIFFERENTIAL EQUATIONS AND LINEAR A

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The initial value problem by Laplace transforms.

Solution Summary: The author explains the Laplace of nth derivative of function, and the Convolution Theorem.

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To solve: The initial value problem by Laplace transforms.

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Chapter 10 Solutions

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The initial value problem by Laplace transforms.