Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises - Page 626: 2

Answer

$2{y^4} = 5{x^2} + C$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{x}{{{y^4}}} \cr & {\text{Separating the variables}} \cr & {y^4}dy = xdx \cr & {\text{Integrate both sides of the equation}} \cr & \int {{y^4}dy} = \int x dx \cr & \frac{{{y^5}}}{5} = \frac{{{x^2}}}{2} + k \cr & {\text{Multiply both sides by 10}} \cr & 2{y^4} = 5{x^2} + 10k \cr & {\text{Let 10}}k = C \cr & 2{y^4} = 5{x^2} + C \cr} $$
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