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

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