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: 18

Answer

$$y=\frac{1}{4}(100x^2+20Cx+C^2)$$

Work Step by Step

$$\frac{dy}{dx}=10\sqrt y$$ Rearranging gives, $$\frac{dy}{\sqrt y}=10dx$$ Integrating both sides. $$\int y^{-1/2}dy=\int 10dx$$ $$\frac{y^{1/2}}{1/2}=10x+c$$ Or, $$y=\frac{1}{4}(100x^2+20Cx+C^2)$$
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