Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises: 2

Answer

$y = (x^2/4 + C_2)^2$

Work Step by Step

$dy/dx = x\sqrt y$ The square root can be written as an exponent (1/2). $dy/dx = xy^{1/2}$ Now let's separate y on one side and x on the other side: $ (1/y^{1/2} )dy = x dx $ Now let's integrate both sides: **Remember that $ 1/y^{1/2}$ can be written as $y^{-1/2}$** $ \int y^{-1/2} dy = \int x dx $ $ y^{1/2}/(1/2) = x^2/2 + C_1 $ $2y^{1/2} = x^2/2 + C_1$ $y^{1/2} = x^2/4 + C_1/2 $ Let's replace $C_1/2 = C_2$ $y = (x^2/4 + C_2)^2$
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