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We are given the initial value y(0) = 2. Substituting this value and solving for L{y} gives the following result. 1 s£{y} − y(0) + 4£{v} = 5¹3 S-3

We are given the initial value y(0) = 2. Substituting this value and solving for L{y} gives the following result. 1 s£{y} − y(0) + 4£{v} = 5¹3 S-3

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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We are given the initial value y(0) = 2. Substituting this value and solving for L{y} gives the following result. SL{y} 2 SL{y} − y(0) + 4L{y} L{y}(s + 4) 1 ) + 4£{y} = 5 ² 3 L{y} = L{y} = 1 5- 3 = 1 S-3 + 2 1 (s - 3)(s + 4) 25- + 2 (s - 3)(s + 4) S + 4 1x )
Transcribed Image Text:We are given the initial value y(0) = 2. Substituting this value and solving for L{y} gives the following result. SL{y} 2 SL{y} − y(0) + 4L{y} L{y}(s + 4) 1 ) + 4£{y} = 5 ² 3 L{y} = L{y} = 1 5- 3 = 1 S-3 + 2 1 (s - 3)(s + 4) 25- + 2 (s - 3)(s + 4) S + 4 1x )
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