## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 10 - Differential Equations - 10.1 Solutions of Elementary and Separable Differential Equations - 10.1 Exercises - Page 535: 8

#### Answer

$$y = k{e^{\frac{{{x^3}}}{3}}}$$

#### Work Step by Step

\eqalign{ & \frac{{dy}}{{dx}} = {x^2}y \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{y} = {x^2}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\frac{{dy}}{y}} = \int {{x^2}dx} \cr & {\text{integrating we obtain}} \cr & \ln \left| y \right| = \frac{{{x^3}}}{3} + C \cr & {\text{solving the equation for }}y \cr & {e^{\ln \left| y \right|}} = {e^{\frac{{{x^3}}}{3} + C}} \cr & {\text{use the property }}{e^{m + n}} = {e^m}{e^n}\,\,\,\left( {{\text{see example 5}}} \right) \cr & {e^{\ln \left| y \right|}} = {e^C}{e^{\frac{{{x^3}}}{3}}} \cr & {\text{simplify by using the logarithmic properties}} \cr & y = {e^C}{e^{\frac{{{x^3}}}{3}}} \cr & {\text{replace the constant }}{e^C}{\text{ with }}k \cr & y = k{e^{\frac{{{x^3}}}{3}}} \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {k{e^{\frac{{{x^3}}}{3}}}} \right] \cr & \frac{{dy}}{{dx}} = k{x^2}{e^{\frac{{{x^3}}}{3}}} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = k{x^2}{e^{\frac{{{x^3}}}{3}}}{\text{ and }}y = k{e^{\frac{{{x^3}}}{3}}}{\text{ into }}\frac{{dy}}{{dx}} = {x^2}y \cr & k{x^2}{e^{\frac{{{x^3}}}{3}}} = {x^2}\left( {k{e^{\frac{{{x^3}}}{3}}}} \right) \cr & k{x^2}{e^{\frac{{{x^3}}}{3}}} = k{x^2}{e^{\frac{{{x^3}}}{3}}} \cr & {\text{The general solution is verified}} \cr}

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