Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.3 Exercises: 16

Answer

The particular solution that satisfies the intial condition is $y=(28-x^{\frac{3}{2}} )^{\frac{2}{3}}$

Work Step by Step

Start by integrating the differential equation using separation of variables $\sqrt x +\sqrt y y' = 0$ $\int \sqrt y dy = \int - \sqrt x dx$ $ \frac{2}{3} y^{\frac{3}{2}} = -\frac{2}{3} x^{\frac{3}{2}} +C$ Let $\frac{3}{2} C = C'$ $y^{\frac{3}{2}} = - x^{\frac{3}{2}} +C'$ Use the initial condition to solve for C' $ (\sqrt 9)^3 = -(\sqrt 1)^3 +C'$ Simplify $C'= 28$ Substitute the C' value back into the general solution $ y^{\frac{3}{2} }= -x^{\frac{3}{2}} +28$ $y=(28-x^{\frac{3}{2}} )^{\frac{2}{3}}$
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