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 - Page 421: 31

Answer

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

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = x \cr & {\text{Separate the variables}} \cr & dy = xdx \cr & {\text{Integrate both sides}} \cr & \int {dy} = \int {xdx} \cr & y = \frac{1}{2}{x^2} + C \cr & \cr & {\text{Graph}} \cr} $$
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