Calculus with Applications (10th Edition)

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

Chapter 10 - Differential Equations - Chapter Review - Review Exercises - Page 561: 30

Answer

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

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = \frac{{{e^x} - x}}{{y - 1}} \cr & {\text{Separating variables leads to}} \cr & \left( {y - 1} \right)dy = \left( {{e^x} - x} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {\left( {y - 1} \right)dy} = \int {\left( {{e^x} - x} \right)dx} \cr & {\text{integrating by using the basic intetgration rules}} \cr & \frac{{{y^2}}}{2} - y = {e^x} - \frac{{{x^2}}}{2} + k \cr & {\text{multiply both sides by 2}} \cr & {y^2} - 2y = 2{e^x} - {x^2} + 2k \cr & {y^2} - 2y = 2{e^x} - {x^2} + C \cr} $$
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