The inverse forms of the results in Problem 49 in Exercises 7.1 are s-a {$ { (5-3)² + b²}= and eat cos bt <((-3)² + ²) = sin be. =eat Use the Laplace transform and these inverses to solve the given initial-value problem.
The inverse forms of the results in Problem 49 in Exercises 7.1 are s-a {$ { (5-3)² + b²}= and eat cos bt <((-3)² + ²) = sin be. =eat Use the Laplace transform and these inverses to solve the given initial-value problem.
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
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Section2.1: Tables And Trends
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![The inverse forms of the results in Problem 49 in Exercises 7.1 are
^{²
{2}.
and
s-a
(sa)² + b²)
y(t)
c{₁5-2)
b
=
eat cos bt
= eat sin bt.
(sa)² + b²
Use the Laplace transform and these inverses to solve the given initial-value problem.
y' + y = et cos 3t, y(0) = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbf9c0aa8-f0c2-40a6-a625-b5a0e0c97da9%2F1a086d69-5115-4ee1-a027-75458dca8e82%2F2ephx4d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The inverse forms of the results in Problem 49 in Exercises 7.1 are
^{²
{2}.
and
s-a
(sa)² + b²)
y(t)
c{₁5-2)
b
=
eat cos bt
= eat sin bt.
(sa)² + b²
Use the Laplace transform and these inverses to solve the given initial-value problem.
y' + y = et cos 3t, y(0) = 0
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