Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 1

Answer

$$y = - 2{x^2} + 2{x^3} + C$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = - 4x + 6{x^2} \cr & {\text{Separating variables leads to}} \cr & dy = - 4xdx + 6{x^2}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = - \int {4x} dx + \int {6{x^2}} dx \cr & {\text{integrate using the power rule}} \cr & y = - 4\left( {\frac{{{x^2}}}{2}} \right) + 6\left( {\frac{{{x^3}}}{3}} \right) + C \cr & y = - 2{x^2} + 2{x^3} + C \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - 2{x^2} + 2{x^3} + C} \right] \cr & \frac{{dy}}{{dx}} = - 2\left( {2x} \right) + 2\left( {3{x^2}} \right) + 0 \cr & \frac{{dy}}{{dx}} = - 4x + 6{x^2} \cr} $$
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