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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

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

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
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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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