Calculus: Early Transcendentals 8th Edition

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

Chapter 17 - Review - Exercises - Page 1181: 5

Answer

$y=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)+e^{2.x}$

Work Step by Step

The equation can be written as: $y''-4.y’+5y=e^{2.x}$ First we solve the homogeneous equation associated to the general one: $y''-4.y'+5y=0$ The caracteristic equation is: $m^2-4.m+5=0$ Solving for m: $Δ=(-4)^2-4.1.5=-4$ $m_1=\frac{4+2.i}{2.1}=2+i$ $m_2=\frac{4-2.i}{2.1}=2-i$ By theorem of homogeneous linear equations, we have: $y_h=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)$ Now we’ll find the particular solution to $y''-4.y’+5y=e^{2.x}$ By undetermined coefficients, taking the general possible solution. We’ll take: $y=a.e^{2.x}$ Differentiating, we have: $y'=2.a.e^{2.x}$ $y''=4.a.e^{2.x}$ Now, putting the informations inside (*), we have: $y''-4.y’+5y=4.a.e^{2.x}-4.2.a.e^{2.x}+5.a.e^{2.x}=e^{2.x}$ Then: $a.e^{2.x}=e^{2.x}$ By equality: $a=1$ Then: $y_p=e^{2.x}$ Therefore: $y=y_h+y_p=c_1.e^{2.x}.sin(x)+c_2.e^{2.x}.cos(x)+e^{2.x}$
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