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The general solution of the 1st-order ODE, (x + 3y − 2)y' = é −3(x+3y−2)² Select one: O a. y(x) - y - 1/3(x - 2) is = In(3x + C) - 2 ○ b._y(x) = ln(6x + C) + (2x)/3 O 1 ○ c. y(x) = 3√ [±ln(18(x + C)) + (2 − x)√/3] O d. y(x) = y(x) = [In(6x + C) ± (2-x)] /3

The general solution of the 1st-order ODE, (x + 3y − 2)y' = é −3(x+3y−2)² Select one: O a. y(x) - y - 1/3(x - 2) is = In(3x + C) - 2 ○ b._y(x) = ln(6x + C) + (2x)/3 O 1 ○ c. y(x) = 3√ [±ln(18(x + C)) + (2 − x)√/3] O d. y(x) = y(x) = [In(6x + C) ± (2-x)] /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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The general solution of the 1st-order ODE, (x + 3y - 2)y' = e−3(x+3y−2)² — y − 1/3(x − 2) is Select one: O a. y(x) = ln(3x + C) - 2 ○ b._ y(x) = [+ln(6x + C) + (2 − x)] /3 ○ c. y(x) = 3√3 [+ln(18(x + C)) + (2 − x)√√3] O d. y(x) = [⁄ln(6x + C) ± (2 - x)]/ /3
Transcribed Image Text:The general solution of the 1st-order ODE, (x + 3y - 2)y' = e−3(x+3y−2)² — y − 1/3(x − 2) is Select one: O a. y(x) = ln(3x + C) - 2 ○ b._ y(x) = [+ln(6x + C) + (2 − x)] /3 ○ c. y(x) = 3√3 [+ln(18(x + C)) + (2 − x)√√3] O d. y(x) = [⁄ln(6x + C) ± (2 - x)]/ /3
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