(x − 2y + 2) dx + (2x + y − 6) dy = 0 In((≈ – 2) + (y – 2)*) — 4arctan In((z − 2) + (3 − 2)^)+4arctan 3 ln ((x − 2)² – (y − 2)²) – 4 arctan y-2 X- 2 X 2 y-2 y ln ((x − 2)² – (y − 2)²) – 4 arctan (2-2) – - x (2-2) In ln((z – 2)* +(y− 2)®)+4arctan| =C =C (2-2) ln ((x − 2)² − (y − 2)²) + 4 arctan (-2) - =C In(( − 2) – (y – 2)^)—4arctan (7-2)= с с с C
(x − 2y + 2) dx + (2x + y − 6) dy = 0 In((≈ – 2) + (y – 2)*) — 4arctan In((z − 2) + (3 − 2)^)+4arctan 3 ln ((x − 2)² – (y − 2)²) – 4 arctan y-2 X- 2 X 2 y-2 y ln ((x − 2)² – (y − 2)²) – 4 arctan (2-2) – - x (2-2) In ln((z – 2)* +(y− 2)®)+4arctan| =C =C (2-2) ln ((x − 2)² − (y − 2)²) + 4 arctan (-2) - =C In(( − 2) – (y – 2)^)—4arctan (7-2)= с с с C
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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Transcribed Image Text:(x – 2y + 2) dx + (2x + y − 6) dy = 0
In((z – 2) + (y – 2)) — 4arctan
In((2 – 2) + (y - 2)^)+4arctan (²-2)
3 ln ((x − 2)² – (y − 2)²) – 4 arctan
ln ((x − 2)² – (y − 2)²) — 4 arctan
In((2 – 2) – (y – 2))+4arctan
In((z – 2) – (y - 2)^) –4arctan
y-2
X 2
In((2 – 2) + (y− 2)) +4arctan
y - 2
X- 2
Y
2
x 2
X
Y
2
2
x 2
y-2
Y 2
X- 2
=C
=C
с
=C
= C
C
с
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