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: 38

Answer

$$y = - 5{e^{ - x}} - 5x + 22$$

Work Step by Step

$$\eqalign{ & \frac{{dy}}{{dx}} = 5\left( {{e^{ - x}} - 1} \right);\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 17 \cr & {\text{Separating variables leads to}} \cr & dy = 5\left( {{e^{ - x}} - 1} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {5\left( {{e^{ - x}} - 1} \right)dx} \cr & \int {dy} = \int {\left( {5{e^{ - x}} - 5} \right)dx} \cr & {\text{integrating}} \cr & y = - 5{e^{ - x}} - 5x + C\,\,\,\,\,\left( {\bf{1}} \right) \cr & \cr & {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 17 \cr & y\left( 0 \right) = 17{\text{ implies that }}y = 17{\text{ when }}x = 0 \cr & {\text{substituting these values into }}\left( {\bf{1}} \right) \cr & 17 = - 5{e^{ - 0}} - 5\left( 0 \right) + C \cr & C = 17 + 5 \cr & C = 22 \cr & {\text{substitute }}C = 3{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr & y = - 5{e^{ - x}} - 5x + 22 \cr} $$
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