Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 17 - Section 17.2 - Nonhomogeneous Linear Equations - 17.2 Exercise - Page 1168: 25


$y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$

Work Step by Step

We are given that $y''-3y'+2y=\dfrac{1}{1+e^{-x}}$ The particular solution is: $y_p=u_1 e^x+u_2 e^{2x}$ Here, we have $u'_1=-(\dfrac{e^{-x}}{1+e^{-x}})$ and $u_1=\int \dfrac{-e^{-x}}{1+e^{-x}}=\ln (1+e^{-x})$ Now, $u'_2=\dfrac{e^{x}}{e^{3x}+e^{2x}} $ and $u_2=\ln (\dfrac{e^x+1}{e^x}-e^{-x})=\ln (1+e^{-x})-e^{x} $ Thus, the particular solution becomes: $y_p=e^x \ln (1+e^{-x}) +e^{2x} [\ln (1+e^{-x})-e^{x}] $ Hence, we get $y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$
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