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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3688a312-1b6e-40d4-973f-537807921925%2F3ee2f329-3032-4e64-b4e8-b45fcb319f77%2Fyh0cqc_processed.png&w=3840&q=75)
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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