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Solving by variation of parameters 4y" 4y'- 8y 8e-*, we find that: O u_1=(-2/9) e^(-3x)and u_2=-(2/3) x O None of them O u_1=(-2/9) e^(-3x) and u_2=x^2/2 O u_1=(-2/9) e^(-3x) and u_2=(-1)/x

Solving by variation of parameters 4y" 4y'- 8y 8e-*, we find that: O u_1=(-2/9) e^(-3x)and u_2=-(2/3) x O None of them O u_1=(-2/9) e^(-3x) and u_2=x^2/2 O u_1=(-2/9) e^(-3x) and u_2=(-1)/x

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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None of them c_1 e^x cosx +c_2 e^x sin x+In(cosx )+x Solving by variation of parameters 4y" - 4y'- 8y 8e-*, we find that: O u_1=(-2/9) e^(-3x)and u_2=-(2/3) x None of them O u_1=(-2/9) e^(-3x) and u_2=x^2/2 O u_1=(-2/9) e^(-3x) and u_2=(-1)/x Using the method of undetermined coefficients, a particular solution of the differential equation y4) – y" = 4x + 2e- is: O (-2/3) x-e^(-x)
Transcribed Image Text:None of them c_1 e^x cosx +c_2 e^x sin x+In(cosx )+x Solving by variation of parameters 4y" - 4y'- 8y 8e-*, we find that: O u_1=(-2/9) e^(-3x)and u_2=-(2/3) x None of them O u_1=(-2/9) e^(-3x) and u_2=x^2/2 O u_1=(-2/9) e^(-3x) and u_2=(-1)/x Using the method of undetermined coefficients, a particular solution of the differential equation y4) – y" = 4x + 2e- is: O (-2/3) x-e^(-x)
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