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(x – 2y + 5)dx – [2(x – 2y) + 9]dy = 0 (x² + y²)(xdy + ydx) – xy(xdx + ydy) = 0 (6x – 3y +2 )dx - (2x – y - 1 )dy = 0 2(x - y)dx + dy = 0

(x – 2y + 5)dx – [2(x – 2y) + 9]dy = 0 (x² + y²)(xdy + ydx) – xy(xdx + ydy) = 0 (6x – 3y +2 )dx - (2x – y - 1 )dy = 0 2(x - y)dx + dy = 0

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 13RE
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For the following equations, identify the substitution equations and solve the solution for each. (Differential Equation) 1 and 4 only
(х — 2у + 5)dx — [2(х — 2у) + 9]dy 3D0 (x? + уxdy + ydх) — ху(xdx + ydy) -0 (6х — Зу +2 )dx - (2х — у - 1 )dy %3Dо 2(х — у)dx + dy %3D0
Transcribed Image Text:(х — 2у + 5)dx — [2(х — 2у) + 9]dy 3D0 (x? + уxdy + ydх) — ху(xdx + ydy) -0 (6х — Зу +2 )dx - (2х — у - 1 )dy %3Dо 2(х — у)dx + dy %3D0
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