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 1167: 4

Answer

$y=y_h+y_p=c_1.e^{x}.cosx+c_2.e^{x}.sinx+\frac{1}{2}.x+\frac{1}{2}+e^{x}$

Work Step by Step

We need to find the general solution of: $y''-2.y+2.y=x+e^{x}$ First, we solve the homogeneous equation associated to the general one: $y''-2.y'+2y=0$ The caracteristic equation is: $m^2-2.m+2=0$ Solving for m: $Δ=(-2)^2-4.1.2=4-8=-4$ $m_1=\frac{(2+\sqrt{(-4)}}{2.1}=\frac{(2+2i)}{2.1}=1+i$ $m_2=\frac{(2-\sqrt{(-4)}}{2.1}=\frac{(2-2i)}{2.1}=1-i$ By theorem of homogeneous linear equations, we have: $y_h=c_1.e^{x}.cosx+c_2.e^{x}.senx$ Now we’ll find the particular solution to $y''-2.y'+2y=x+e^{x} (*)$ Taking the general possible solution: $y=(a.x+b)+c.e^{x}$ Differentiating, we have: $y'=a+c.e^{x}$ $y''=c.e^{x}$ Now, putting the informations inside (*), we have: $y''-2.y'+2y=c.e^{x}-2.(a+c.e^{x})+2.[(a.x+b)+c.e^{x}]=1.x+1.e^{x}$ Then: $c.e^{x}-2.a-2.c.e^{x}+2.a.x+2.b+2.c.e^{x}=1.x+1.e^{x} (-2.a+2.b)+c.e^{x}+2.a.x=1.x+1.e^{x}$ By equality: $-2.a+2.b=0$ $c=1$ $2.a=1$ By solving the equations, we have: $c=1$ $a=\frac{1}{2}$ $-2.\frac{1}{2} + 2.b = 0$ $-1+2.b=0$ $2.b = 1$ $b=\frac{1}{2}$ Then $y_p=\frac{1}{2}.x+\frac{1}{2}+e^{x}$ Therefore, the general solution is: $y=y_h+y_p=c_1.e^{x}.cosx+c_2.e^{x}.sinx+\frac{1}{2}.x+\frac{1}{2}+e^{x}$
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