## 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: 30

#### Answer

$$y = - \ln \left( {\frac{{{{\left( {x + 2} \right)}^3} - 30}}{3}} \right)$$

#### Work Step by Step

\eqalign{ & \frac{{dy}}{{dx}} = {\left( {x + 2} \right)^2}{e^y};\,\,\,\,\,\,\,\,\,\,\,y\left( 1 \right) = 0 \cr & {\text{Separating variables leads to}} \cr & \frac{{dy}}{{{e^y}}} = {\left( {x + 2} \right)^2}dx \cr & {\text{use }}\frac{1}{{{u^n}}} = {u^{ - n}} \cr & {e^{ - y}}dy = {\left( {x + 2} \right)^2}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {{e^{ - y}}dy} = \int {{{\left( {x + 2} \right)}^2}dx} \cr & {\text{integrate by using the power rule for integration and }}\int {{e^u}du} = {e^u} + C \cr & - {e^{ - y}} = \frac{{{{\left( {x + 2} \right)}^3}}}{3} + C\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,\,y\left( 1 \right) = 0 \cr & \,\,y\left( 1 \right) = 0{\text{ implies that }}y = 0{\text{ when }}x = 1 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & - {e^{ - 0}} = \frac{{{{\left( {1 + 2} \right)}^3}}}{3} + C \cr & - 1 = 9 + C \cr & C = - 10 \cr & {\text{substitute }}C = - 10{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & - {e^{ - y}} = \frac{{{{\left( {x + 2} \right)}^3}}}{3} - 10 \cr & {\text{Solve the equation for }}y \cr & - {e^{ - y}} = 10 - \frac{{{{\left( {x + 2} \right)}^3}}}{3} \cr & - {e^{ - y}} = \frac{{30 - {{\left( {x + 2} \right)}^3}}}{3} \cr & {e^{ - y}} = \frac{{{{\left( {x + 2} \right)}^3} - 30}}{3} \cr & \ln {e^{ - y}} = \ln \left( {\frac{{{{\left( {x + 2} \right)}^3} - 30}}{3}} \right) \cr & y = - \ln \left( {\frac{{{{\left( {x + 2} \right)}^3} - 30}}{3}} \right) \cr}

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