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Solve (x – y)dx + (3x + y)dy = 0 with the condition: y(2) = –1 O (2æ + 4y) – (2x + 4y) ln(2x + 4y) = 0 O (2x + 4y) – (x+ y) ln(x + y) = 2 O (x + 2y) + (x+ y) ln(x + y) = 2 O 2 (x + 2y) + (x + y) In(x + y) = 0

Solve (x – y)dx + (3x + y)dy = 0 with the condition: y(2) = –1 O (2æ + 4y) – (2x + 4y) ln(2x + 4y) = 0 O (2x + 4y) – (x+ y) ln(x + y) = 2 O (x + 2y) + (x+ y) ln(x + y) = 2 O 2 (x + 2y) + (x + y) In(x + y) = 0

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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Solve (х — у)dx + (3x + y)dy — 0 with the condition: y(2) = -1 O (2x + 4y) – (2x + 4y) In(2x + 4y) = 0 O (2x + 4y) – (x + y) In(x + y) = 2 O (x + 2y) + (x + y) ln(x + y) = 2 о 2 ( + 2у) + (т + у) ln(x + y) %3D0
Transcribed Image Text:Solve (х — у)dx + (3x + y)dy — 0 with the condition: y(2) = -1 O (2x + 4y) – (2x + 4y) In(2x + 4y) = 0 O (2x + 4y) – (x + y) In(x + y) = 2 O (x + 2y) + (x + y) ln(x + y) = 2 о 2 ( + 2у) + (т + у) ln(x + y) %3D0
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