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: 11

Answer

y=$-\frac{1}{4}\sqrt {1-4x^{2}}$+C

Work Step by Step

$\sqrt {1-4x^{2}}y'$=x dy=$\frac{x}{\sqrt {1-4x^{2}}}$dx $\int$dy=$\int$$\frac{x}{\sqrt {1-4x^{2}}}$dx $\int$dy=$\frac{-1}{8}$$\int$(1-4$x^{2}$)$^{-\frac{1}{2}}$(-8xdx) y=$-\frac{1}{4}\sqrt {1-4x^{2}}$+C
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