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: 14


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

Work Step by Step

Given: $\sqrt {2yx}\dfrac{dy}{dx}=1$ Re-arrange the given equation and inetgrate as follows:. Then $ \int \dfrac{dx}{\sqrt {2x}}=\int \sqrt y dy$ ....(1) As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$ Equation (1) becomes: $ \sqrt 2 \int \dfrac{dx}{2 \sqrt {x}}=\int y^{1/2} dy \implies \dfrac{y^{1/2+1}}{1/2+1}=\sqrt {2x}+c$ Hence, $\dfrac{2}{3}y^{3/2}-\sqrt {2x}=c$
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