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(2xy- 3x)dx+(x² + 2y)dy = 0

(2xy- 3x)dx+(x² + 2y)dy = 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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please look at the sample computation and compute the differential equation in the same way

(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
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
(1) (2xy – 3x²)dx + (x² + 2y)dy = 0
Transcribed Image Text:(1) (2xy – 3x²)dx + (x² + 2y)dy = 0
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