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: 5

Answer

$${y^2} = \frac{{2{x^3}}}{3} + C$$

Work Step by Step

$$\eqalign{ & y\frac{{dy}}{{dx}} = {x^2} \cr & {\text{Separating variables leads to}} \cr & ydy = {x^2}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {ydy} = \int {{x^2}} dx \cr & {\text{integrate by using the power rule}} \cr & \frac{{{y^2}}}{2} = \frac{{{x^3}}}{3} + C \cr & {\text{multiplying both sides by 2}} \cr & \boxed{{y^2} = \frac{{2{x^3}}}{3} + C} \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{d}{{dx}}\left[ {{y^2}} \right] = \frac{d}{{dx}}\left[ {\frac{{2{x^3}}}{3} + C} \right] \cr & 2y\frac{{dy}}{{dx}} = 2{x^2} \cr & \frac{{dy}}{{dx}} = \frac{{{x^2}}}{y} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = \frac{{{x^2}}}{y}{\text{ into }}y\frac{{dy}}{{dx}} = {x^2} \cr & y\left( {\frac{{{x^2}}}{y}} \right) = {x^2} \cr & {x^2} = {x^2} \cr & {\text{The general solution is verified}} \cr} $$
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