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4, xydx (x² + 3y²) dy = 0 Homogeneous, 2nd degree Let x = vy dx= vdy + ydv Substitute. vy² (vdy+ydv)-(v²y2 + 3y2) dy=0 vy² (vdy + ydv)- y² (v²+3) dy = 0. divide by y2, v²dy + vydv (v²+3) dy = 0 vydv - 3dy = 0 divide by y. vdv - but : 3 dy y fvdv - 3 fo ²-3 ln y = -3 Inc v²-6lny = -6 Inc y² = 6lny - 6 Inc v² = = 0 6 In v = ²=6 = 6 In y/c x² = 6y² In y/c ans.

4, xydx (x² + 3y²) dy = 0 Homogeneous, 2nd degree Let x = vy dx= vdy + ydv Substitute. vy² (vdy+ydv)-(v²y2 + 3y2) dy=0 vy² (vdy + ydv)- y² (v²+3) dy = 0. divide by y2, v²dy + vydv (v²+3) dy = 0 vydv - 3dy = 0 divide by y. vdv - but : 3 dy y fvdv - 3 fo ²-3 ln y = -3 Inc v²-6lny = -6 Inc y² = 6lny - 6 Inc v² = = 0 6 In v = ²=6 = 6 In y/c x² = 6y² In y/c ans.

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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where does the -3ln C comes from? 

4, xydx (x² + 3y²) dy = 0 Homogeneous, 2nd degree Let x = vy dx= vdy + ydv Substitute, vy² (vdy+ydv)-(v²y2 + 3y2) dy=O vy² (vdy + ydv) - y² (v²+ 3) dy = 0. divide by y2, v²dy + vydv (v²+3) dy = 0 vy dv 3 dy = 0 - divide by y. vdv - 3 dy y = 0 fvdv3fd-fo ²-3 ln y = -3 in C v²-6lny = -6 Inc y² = 6lny - 6 In c v² = 6 In but : V = = = 6 in 9/c x² = 6y² In y/c ans.
Transcribed Image Text:4, xydx (x² + 3y²) dy = 0 Homogeneous, 2nd degree Let x = vy dx= vdy + ydv Substitute, vy² (vdy+ydv)-(v²y2 + 3y2) dy=O vy² (vdy + ydv) - y² (v²+ 3) dy = 0. divide by y2, v²dy + vydv (v²+3) dy = 0 vy dv 3 dy = 0 - divide by y. vdv - 3 dy y = 0 fvdv3fd-fo ²-3 ln y = -3 in C v²-6lny = -6 Inc y² = 6lny - 6 In c v² = 6 In but : V = = = 6 in 9/c x² = 6y² In y/c ans.
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