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Solving by variation of parameters y"-3y =x we find that: O None of them O u1=(-1/6) x, u_2=-e^(-3x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-2x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-3x) (x/9+1/27)

Solving by variation of parameters y"-3y =x we find that: O None of them O u1=(-1/6) x, u_2=-e^(-3x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-2x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-3x) (x/9+1/27)

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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Solving by variation of parameters y" - 3y' = x we find that: O None of them O u_1=(-1/6) x, u_2=-e^(-3x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-2x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-3x) (x/9+1/27)
Transcribed Image Text:Solving by variation of parameters y" - 3y' = x we find that: O None of them O u_1=(-1/6) x, u_2=-e^(-3x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-2x) (x/9+1/27) u_1=(-1/6) x^2, u_2=-e^(-3x) (x/9+1/27)
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