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If y" – 2y' + 3y = 0, y(0) = -2, y'(0) = 1, then L{y(t)} = Y(s) = a) 2s+3 82–2s+3 b) 2s+5 g2–2s+3 c) -2s-3 s2-2s+3

If y" – 2y' + 3y = 0, y(0) = -2, y'(0) = 1, then L{y(t)} = Y(s) = a) 2s+3 82–2s+3 b) 2s+5 g2–2s+3 c) -2s-3 s2-2s+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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If y" – 2y' + 3y = 0, y(0) = -2, y'(0) = 1, then L{y(t)} = Y(s) = a) 2s+3 s2-2s+3 2s+5 b) s2-2s+3 c) -2s-3 s2-2s+3 d) None of these e) -2s+5 s2-2s+3
Transcribed Image Text:If y" – 2y' + 3y = 0, y(0) = -2, y'(0) = 1, then L{y(t)} = Y(s) = a) 2s+3 s2-2s+3 2s+5 b) s2-2s+3 c) -2s-3 s2-2s+3 d) None of these e) -2s+5 s2-2s+3
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