Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 17 - Second-Order Differential Equations - 17.2 Nonhomogeneous Linear Equations - 17.2 Exercises - Page 1208: 28

Answer

$y=c_1e^{-2x}+c_2 xe^{-2x}+\dfrac{e^{-2x}}{2x} $

Work Step by Step

The auxiliary solution is $r^2+4r+1=0 \implies r=-2$ Here, $y_c=c_1 e^{-2x}+c_2 xe^{-2x}$ $y_p=u_1 e^{-2x}+u_2xe^{-2x}; y'_p=u'_1 e^{-2x}-2u_1 e^{-2x}+u'_2 xe^{-2x}+u_2 e^{-2x}(1-2x)$ This gives: $u'_1=--u'_2 x$ Here, we have $y''+4y'+4y=\dfrac{e^{-2x}}{x^3}$ and $-2u'_1+u'_2(1-2x)=\dfrac{1}{x^3}$ $u'_2=\dfrac{1}{x^3} $ or, $u_2=\dfrac{-1}{2x^2}$ and $u'_1=-[\dfrac{1}{x^2}]$ or, $u_1=x^{-1}$ Thus, $y_p=u_1 e^{-2x}+u_2xe^{-2x} $ or, $y=\dfrac{e^{-2x}}{x} -\dfrac{xe^{-2x}}{2x^2} =\dfrac{e^{-2x}}{2x} $ Hence, we have $y=y_c+y_p$ or, $y=c_1e^{-2x}+c_2 xe^{-2x}+\dfrac{e^{-2x}}{2x} $
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