Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - Review Exercises - Page 432: 34



Work Step by Step

$\frac{dy}{dx}=\frac{x^3}{2y^2}$ Therefore, $2y^2dy=x^3dx$ $\int 2y^2dy=\int x^3dx$ Or, $\frac{2y^3}{3}=\frac{x^4}{4}+C'$ Multiplying both sides by 12 and taking $C=12C'$. $$8y^3=3x^4+C$$
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