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: 17


$y=\sin (x^2+c)$

Work Step by Step

Given: $\dfrac{dy}{dx}=2x \sqrt {1-y^2}$ Re-arrange the given equation and integrate as follows:. Then $\int(2x)dx= \int \dfrac{dy}{\sqrt {1-y^2}}$....(1) As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$ Equation (1) becomes: $ \sin^{-1} y=x^2+c$ Now, consider taking $\sin$ of both sides. Then $\sin (\sin^{-1} y)=\sin (x^2+c)$ Hence, $y=\sin (x^2+c)$
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