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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Question
please look at the sample computation and compute the differential equation in the same way
![(3) xexydx + ye*Ydy = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a54d7b5-e336-45e0-8ae1-38c29924b6d9%2F53f3270d-5fcb-4f7f-b94a-69f30de3ab0b%2Fhaf5f7s_processed.png&w=3840&q=75)
Transcribed Image Text:(3) xexydx + ye*Ydy = 0
![(y2
- 2xy + 6x)dx – (x² – 2xy + 2)dy = 0
(y² –
- 2xy + 6x)dx +(-x² + 2xy– 2)dy = 0
Testing for exactness:
Treating y as a constant,
ƏM
a
(y2
- 2xy + 6x) = 2y – 2x
F= y* f dx – 2y f xdx + 6 S xdx + ¢O°
F = v?
ду
ду
ƏN
(-x² + 2xy - 2) = -2x + 2y = 2y - 2x
ax
F = y2(x) – 2y
+ 6
+ ¢(y)
ƏF
= M
ax
Our DE is exact. Now we take
F = xy? – x²y + 3x² + $(v) (1)
S ar = f
Mdx + $(y)
F =
- 2xy + 6x)dx + $(y)
F = xy? – x²y + 3x² + ¢(y) (1)
Solving for its integral:
Jø'w) = [ -2dy
Now taking its partial derivative with respect to y:
ƏF
$(v) = -2y
%3D
;(xy² – x²y + 3x² + ¢(y))
ду ду
Substituting it to (1),
2xy – x² + o'(y)
F = xy? – x²y + 3x² – 2y
ду
Our solution is:
But:
xy² – x²y + 3x2 – 2y = c
ƏF
= N = -x2 + 2xy - 2
ду
Then
ду
ду
2xy – x² + o'(y) = -x² + 2xy – 2
p'(y) = -2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a54d7b5-e336-45e0-8ae1-38c29924b6d9%2F53f3270d-5fcb-4f7f-b94a-69f30de3ab0b%2Fl28h3dh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(y2
- 2xy + 6x)dx – (x² – 2xy + 2)dy = 0
(y² –
- 2xy + 6x)dx +(-x² + 2xy– 2)dy = 0
Testing for exactness:
Treating y as a constant,
ƏM
a
(y2
- 2xy + 6x) = 2y – 2x
F= y* f dx – 2y f xdx + 6 S xdx + ¢O°
F = v?
ду
ду
ƏN
(-x² + 2xy - 2) = -2x + 2y = 2y - 2x
ax
F = y2(x) – 2y
+ 6
+ ¢(y)
ƏF
= M
ax
Our DE is exact. Now we take
F = xy? – x²y + 3x² + $(v) (1)
S ar = f
Mdx + $(y)
F =
- 2xy + 6x)dx + $(y)
F = xy? – x²y + 3x² + ¢(y) (1)
Solving for its integral:
Jø'w) = [ -2dy
Now taking its partial derivative with respect to y:
ƏF
$(v) = -2y
%3D
;(xy² – x²y + 3x² + ¢(y))
ду ду
Substituting it to (1),
2xy – x² + o'(y)
F = xy? – x²y + 3x² – 2y
ду
Our solution is:
But:
xy² – x²y + 3x2 – 2y = c
ƏF
= N = -x2 + 2xy - 2
ду
Then
ду
ду
2xy – x² + o'(y) = -x² + 2xy – 2
p'(y) = -2
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