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Given the differential equation y" +3y + 2y =D Which of the (1+e²) following equations result after the proposed particular solution is substituted into the given differential equation. ex -2x 1+ ex -2c,(x)e-2* – c,(x)e¬*= 0 1 2x %3D 1+ex -2c,(x)e-2x – c,(x)e¬*= 0 o c)e-2x + c,(x)e-*= 0 1 -c(x)e-2x - 2c,(x)e¯* = 1+ ex oc(x)e-2* +c,(x)e-*= 0 1 -2x %3D 1+ex

Given the differential equation y" +3y + 2y =D Which of the (1+e²) following equations result after the proposed particular solution is substituted into the given differential equation. ex -2x 1+ ex -2c,(x)e-2* – c,(x)e¬*= 0 1 2x %3D 1+ex -2c,(x)e-2x – c,(x)e¬*= 0 o c)e-2x + c,(x)e-*= 0 1 -c(x)e-2x - 2c,(x)e¯* = 1+ ex oc(x)e-2* +c,(x)e-*= 0 1 -2x %3D 1+ex

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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Given the differential equation y " +3y + 2y% = Which of the (1+e)· following equations result after the proposed particular solution is substituted into the given differential equation. ex c(x)e-2x + c,(x)e¬x = 1+ex -2c,(x)e-2* - c;(x)e¯* =0 c,(x)e-2* + c¿(x)e¬* =. 1 + e* 1 -2c,(x)e-2x - c,(x)e¯x= 0 O c(x)e-2*+ c,(x)e¯* 0 1 -c;(x)e-2x – 2c,(x)e¯x= 1+ex O c(x)e-2X+c,(x)e-*= 0 -2c(x)e-2* - c(x)e-x=! mpus 1+ex
Transcribed Image Text:Given the differential equation y " +3y + 2y% = Which of the (1+e)· following equations result after the proposed particular solution is substituted into the given differential equation. ex c(x)e-2x + c,(x)e¬x = 1+ex -2c,(x)e-2* - c;(x)e¯* =0 c,(x)e-2* + c¿(x)e¬* =. 1 + e* 1 -2c,(x)e-2x - c,(x)e¯x= 0 O c(x)e-2*+ c,(x)e¯* 0 1 -c;(x)e-2x – 2c,(x)e¯x= 1+ex O c(x)e-2X+c,(x)e-*= 0 -2c(x)e-2* - c(x)e-x=! mpus 1+ex
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