а. (2xy² – y)dx + xdy = 0 2xy² dy b. . dx 2x2у + 1 c. (x – 1)²ydx + x²(y + 1)dy = 0 - (1+ e*y+xe*y)dx + (xe* + 2)dy = 0

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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Which is the homogeneous equation?

а.
(2xy² – y)dx + xdy = 0
dy
2ху?
b. .
dx
2x²y + 1
c. (x – 1)²ydx + x²(y + 1)dy = 0
d. (1+ e*y+xe*y)dx + (xe* + 2)dy = 0
Transcribed Image Text:а. (2xy² – y)dx + xdy = 0 dy 2ху? b. . dx 2x²y + 1 c. (x – 1)²ydx + x²(y + 1)dy = 0 d. (1+ e*y+xe*y)dx + (xe* + 2)dy = 0
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