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(2x + 3y – 1) dx – 4 (x + 1) dy = 0 a.) (2x + y – 3)4 = c (x + 1)3 b.) (2x + y + 3)4 = c (x + 1)3 c.) (2x - y + 3)4 = c (x + 1)3 d.) (2x - y – 3)4 = c (x + 1)3

(2x + 3y – 1) dx – 4 (x + 1) dy = 0 a.) (2x + y – 3)4 = c (x + 1)3 b.) (2x + y + 3)4 = c (x + 1)3 c.) (2x - y + 3)4 = c (x + 1)3 d.) (2x - y – 3)4 = c (x + 1)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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What is the solution for this DE. 

(2х + Зу — 1) dx — 4 (x + 1) dy 3D0 а.) (2х + у - 3)4%3D с (x + 1)3 b.) (2x + у + 3)4%3D с (х + 1)3 с.) (2х - у + 3)4%3D с (х + 1)3 d.) (2x - у — 3)4 %3D с (x + 1)3
Transcribed Image Text:(2х + Зу — 1) dx — 4 (x + 1) dy 3D0 а.) (2х + у - 3)4%3D с (x + 1)3 b.) (2x + у + 3)4%3D с (х + 1)3 с.) (2х - у + 3)4%3D с (х + 1)3 d.) (2x - у — 3)4 %3D с (x + 1)3
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