The inverse forms of the results in Problem 49 in Exercises 7.1 are [{5-²0 and s-a y(t) = (sa)² + b² = e cos bt b * ½{15- { (5-2)² +6² } = eat sin bt. Use the Laplace transform and these inverses to solve the given initial-value problem. y' + y = e-4t cos 3t, y(0) = 0

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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The inverse forms of the results in Problem 49 in Exercises 7.1 are
and
s-a
2³
y(t) =
(sa)² + b²
b
*{{(8-2) + b²}
(sa)² + b²
= e
cos bt
= eat sin bt.
Use the Laplace transform and these inverses to solve the given initial-value problem.
y' + y = e-4t cos 3t, y(0) = 0
Transcribed Image Text:The inverse forms of the results in Problem 49 in Exercises 7.1 are and s-a 2³ y(t) = (sa)² + b² b *{{(8-2) + b²} (sa)² + b² = e cos bt = eat sin bt. Use the Laplace transform and these inverses to solve the given initial-value problem. y' + y = e-4t cos 3t, y(0) = 0
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