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Evaluate y" - y' - 2y = sin2x О а. у%3 (С1* e^-x) + (C2 * е^2х) - (3/20 * sin2x) + (1/20 * сos2x) оБ. у%3 (С1*e^-x) + (С2 * е^2х) - (3/20 * sin2x) + (3/20 * сos2x) None of the choices O d. y%3D(С1* e^2x) + (C2 * е^-х) - (1/20 * sin2x) + (1/20 * сos2x) Ое. у%3 (С1* е^2х) + (С2* е^-х) - (1/20 * sin2x) + (3/20 * сos2x)

Evaluate y" - y' - 2y = sin2x О а. у%3 (С1* e^-x) + (C2 * е^2х) - (3/20 * sin2x) + (1/20 * сos2x) оБ. у%3 (С1*e^-x) + (С2 * е^2х) - (3/20 * sin2x) + (3/20 * сos2x) None of the choices O d. y%3D(С1* e^2x) + (C2 * е^-х) - (1/20 * sin2x) + (1/20 * сos2x) Ое. у%3 (С1* е^2х) + (С2* е^-х) - (1/20 * sin2x) + (3/20 * сos2x)

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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Evaluate y" - y' - 2y = sin2x O a. y= (C1 * e^-x) + (C2 * e^2x) - (3/20 * sin2x) + (1/20 * cos2x) b. y = (C1 * e^-x) + (C2 * e^2x) - (3/20 * sin2x) + (3/20 * cos2x) C. None of the choices O d. y = (C1 * e^2x) + (C2 * e^-x) - (1/20 * sin2x) + (1/20 * cos2x) O e. y = (C1 * e^2x) + (C2 * e^-x) - (1/20 * sin2x) + (3/20 * cos2x)
Transcribed Image Text:Evaluate y" - y' - 2y = sin2x O a. y= (C1 * e^-x) + (C2 * e^2x) - (3/20 * sin2x) + (1/20 * cos2x) b. y = (C1 * e^-x) + (C2 * e^2x) - (3/20 * sin2x) + (3/20 * cos2x) C. None of the choices O d. y = (C1 * e^2x) + (C2 * e^-x) - (1/20 * sin2x) + (1/20 * cos2x) O e. y = (C1 * e^2x) + (C2 * e^-x) - (1/20 * sin2x) + (3/20 * cos2x)
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