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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Chapter2: Graphical And Tabular Analysis
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
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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