Explanation Given: The initial value problem is, d 2 y d t 2 + 4 d y d t + 13 y = 10 sin 5 t ... ( 1 ) And the initial conditions are y ( 0 ) = 0 , and d y d t ( 0 ) = 0 . Approach: The Laplace of n th derivative of the function L { f n ( t ) } exists for S > a and 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 ) The Laplace transform of L [ δ ( t − a ) ] is given by, L [ δ ( t − a ) ] = e − s a ... ( 2 ) The Laplace inverse of 1 s − a is given by, L − 1 { 1 s − a } = e a t ... ( 3 ) Calculation: Consider equation ( 1 ) and impulsive force on the mass at t = 10 in the downward positive direction obtains the initial value problem, y ″ + 4 y + 13 y = 10 sin 5 t + 2 δ ( t − 10 ) Take Laplace Transform both sides of above yields, [ s 2 Y ( s ) − s y ( 0 ) − y ′ ( 0 ) ] + 4 [ s Y ( s ) − y ( 0 ) ] + 13 Y ( s ) = 50 s 2 + 25 + 2 e − 10 s Substitute 0 for y ( 0 ) and 0 for y ′ ( 0 ) in above. [ s 2 Y ( s ) ] + 4 [ s Y ( s ) ] + 13 Y ( s ) = 50 s 2 + 25 + 2 e − 10 s Y ( s ) ( s 2 + 4 s + 13 ) = 50 s 2 + 25 + 2 e − 10 s It implies, Y ( s ) = 50 ( s 2 + 25 ) ( ( s + 2 ) 2 + 9 ) + 2 e − 10 s ( s + 2 ) 2 + 9 = 1 68 [ 175 ( s + 2 ) 2 + 9 + 25 s ( ( s + 2 ) 2 + 9 ) − 15 ( 5 s 2 + 25 ) − 25 ( s s 2 + 25 ) ] + 2 e − 10 s ( s + 2 ) 2 + 9 = 1 68 [ 175 ( s + 2 ) 2 + 9 + 25 ( s + 2 ( ( s + 2 ) 2 + 9 ) − 2 ( s + 2 ) 2 + 9 ) − 15 ( 5 s 2 + 25 ) − 25 ( s s 2 + 25 ) ] + 2 e − 10 s ( s + 2 ) 2 + 9 Take the inverse Laplace Transform both sides to obtain

Differential Equations and Linear Algebra (4th Edition)

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Author: Stephen W. Goode, Scott A. Annin

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Chapter 10.8, Problem 15P

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The resulting motion of the mass for the initial value problem is, d2ydt2+4dydt+13y=10sin5t and the initial conditions are y(0)=0, and dydt(0)=0.

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Chapter 10.8, Problem 14P

Chapter 10.8, Problem 16P

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The resulting motion of the mass for the initial value problem is, d 2 y d t 2 + 4 d y d t + 13 y = 10 sin 5 t and the initial conditions are y ( 0 ) = 0 , and d y d t ( 0 ) = 0 .

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The resulting motion of the mass for the initial value problem is, d2ydt2+4dydt+13y=10sin5t and the initial conditions are y(0)=0, and dydt(0)=0.

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