Explanation Given: The initial value problem is, y ′ ′ + ω 2 y = A sin ω 0 t + B cos ω 0 t ... ( 1 ) The initial conditions are, y ( 0 ) = y 0 , y ′ ( 0 ) = y 1 , Approach: The Laplace transform of n th derivative of a 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 ) ... ( 2 ) The Laplace of [ cos b t ] function is given by, L { t } = s s 2 + b 2 ... ( 3 ) The Laplace of [ sin b t ] function is given by, L { t } = b s 2 + b 2 ... ( 4 ) Calculation: Take the Laplace transform of equation ( 1 ) . L ( Y ′ ′ ) + ω 2 L [ Y ] = L [ A sin ω 0 t + B cos ω 0 t ] ... ( 5 ) From equation ( 2 ) , Laplace of L ( Y ′ ′ ) is, L ( Y ′ ′ ) = s 2 Y ( s ) − s Y ( 0 ) − Y ′ ( 0 ) From equation ( 3 ) , Laplace of L ( A sin ω 0 t + B cos ω 0 t ) is, L ( A sin ω 0 t + B cos ω 0 t , ) = A ω 0 s 2 + ω 0 2 + B s s 2 + ω 0 2 So, equation ( 5 ) can be written as, { s 2 Y ( s ) − s Y ( 0 ) − Y ′ ( 0 ) } + ω 2 Y ( s ) = A ω 0 s 2 + ω 0 2 + B s s 2 + ω 0 2 ( s 2 + ω 2 ) Y ( s ) − s y 0 − y 1 = A ω 0 s 2 + ω 0 2 + B s s 2 + ω 0 2 Y ( s ) = A ω 0 ( s 2 + ω 2 ) ( s 2 + ω 0 2 ) + B s ( s 2 + ω 2 ) ( s 2 + ω 0 2 ) + s y 0 ( s 2 + ω 2 ) + y 1 ( s 2 + ω 2 ) Suppose, A ω 0 + B s ( s 2 + ω 2 ) ( s 2 + ω 0 2 ) = P s − Q s 2 + ω 2 + R s + T s 2 + ω 0 2 = ( P + R ) s 3 + ( T − Q ) s 2 + ( P ω 0 2 + R ω 2 ) + ( T ω 2 − Q ω 0 2 ) ( s 2 + ω 2 ) ( s 2 + ω 0 2 ) ... ( 6 ) Compare the coefficients on both sides of the above equation. P + R = 0 T − Q = 0 P ω 0 2 + R ω 2 = B ω 0 T ω 2 − Q ω 0 2 = A Here, − P = R = B ω 0 ω 2 − ω 0 2 and T = Q = A ω 2 − ω 0 2 Substitute − B ω 0 ω 2 − ω 0 2 for P , B ω 0 ω 2 − ω 0 2 for R , A ω 2 − ω 0 2 for T and A ω 2 − ω 0 2 for Q in equation ( 6 )

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ISBN: 9780321990167

Author: ANNIN

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Chapter 10.4, Problem 29P

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The initial value problem y′′+ω2y=Asinω0t+Bcosω0t by Laplace transforms.

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Chapter 10.4, Problem 28P

Chapter 10.4, Problem 30P

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The initial value problem y ′ ′ + ω 2 y = A sin ω 0 t + B cos ω 0 t by Laplace transforms.

Solution Summary: The author explains how to solve the initial value problem y(t) by Laplace transforms.

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To solve:

The initial value problem y′′+ω2y=Asinω0t+Bcosω0t by Laplace transforms.

Want to see the full answer?

Chapter 10 Solutions

The initial value problem y ′ ′ + ω 2 y = A sin ω 0 t + B cos ω 0 t by Laplace transforms.

Solution Summary: The author explains how to solve the initial value problem y(t) by Laplace transforms.

Videos

To solve: The initial value problem y′′+ω2y=Asinω0t+Bcosω0t by Laplace transforms.

Want to see the full answer?

Chapter 10 Solutions

To solve:

The initial value problem y′′+ω2y=Asinω0t+Bcosω0t by Laplace transforms.