University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.2 - Exponential Change and Separable Differential Equations - Exercises - Page 409: 10


$2 \sqrt y -\dfrac{x^3}{3} =C$

Work Step by Step

We have: $\dfrac{dy}{dx}=x^2 \sqrt y$ or, $\dfrac{dy}{\sqrt y}=x^2 dx$ We have to integrate the above expression: $\int \dfrac{dy}{\sqrt y}= \int x^2 dx$ or, $2 (\int \dfrac{dy}{2\sqrt y})= \int x^2 dx$ This implies: $2 \sqrt y=\dfrac{x^3}{3}+C$ Hence, $2 \sqrt y -\dfrac{x^3}{3} =C$
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