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: 9

Answer

$y=2e^x+\frac{1}{2}x^2e^x-xe^x$

Work Step by Step

$y''-y'=xe^x$ $r^2-r=0$ $r(r-1)=0$ $r=0$ v $r=1$ $y=c_{1}+c_{2}e^x$ $y_{p}=x(Ax+B)e^x=Ax^2+Bxe^x$ $y_{p}'=(2Ax+Ax^2)e^x+(Bx+B)e^x$ $y_{p}''=(Ax^2+4Ax+2A)e^x+(Bx+2B)e^x$ $(Ax^2+4Ax+2A)e^x+(Bx+2B)e^x+(Bx+2B)e^x=(2Ax+2A)e^x+Be^X=xe^x$ $(2A)xe^x+(2A+B)e^x=xe^x$ $A=\frac{1}{2}$ and $B=-1$ $y_{p}=\frac{1}{2}x^2e^x-xe^x$ $y=c_{1}+c_{2}e^x+\frac{1}{2}x^2e^x-xe^x$ $y(0)=2$ which indicates $c_{1}+c_{2}=2$ $y'=0+c_{2}+xe^x+\frac{1}{2}x^2e^x-e^x-xe^x$ $y'(0)=c_{2}+0+0-1+0$ gives $y'(0)=c_{2}-1=1$ $c_{1}=0$ and $c_{2}=2$ $y=2e^x+\frac{1}{2}x^2e^x-xe^x$
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