Calculus 10th Edition

Published by Brooks Cole

Chapter 6 - Differential Equations - 6.1 Exercises: 3

Answer

$x^{2}+y^{2}=Cy$

Work Step by Step

$2x=2yy'=Cy'$ $y'=(-2x)/(2y-C)$ $y'=(-2xy)/(2y^{2}-Cy)$ $y'=(-2xy)/(2y^{2}-(x^{2}+y^{2})$ $y'=(-2xy)/(y^{2}-x^{2})$ $y'=(2xy)/(x^{2}-y^{2})$ $x^{2}+y^{2}=Cy$

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