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 1207: 10

Answer

$y=\frac{1}{6}e^{-2x}+\frac{17}{15}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$

Work Step by Step

$y''+y'-2y=0$ Use auxiliary equation $r^{2}+r-2=0$ $(r+2)(r-1)=0$ $r_{1}=-2$ $r_{2}=1$ $y_{c}=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$ $y_{c}=c_{1}e^{-2x}+c_{2}e^{x}$ The particular solution is the form $y_{p}=A+Bx+Csin2x+Dcos2x$ $y_{p}'=B+2Ccos2x-2Dsin2x$ $y_{c}''=-4Csin2x-4Dcos2x$ Substitute into the main differential equation $(-4Csin2x-4Dcos2x)+(B+2Ccos2x-2Dsin2x)-2(A+Bx+Csin2x+Dcos2x)=x+sin2x$ $(B-2A)-2Bx+sin2x(-4C-2D-2C)+cos2x(-4D+2C-2D)=x+sin2x$ $(B-2A)-2Bx+sin2x(-6C-2D)+cos2x(-6D+2C)=x+sin2x$ $-2B=1$ $B=-\frac{1}{2}$ $B-2A=0$ $A=\frac{B}{2}$ $A=-\frac{1}{4}$ $-6C-2D=1$ $2C-6D=0$ Multiply both equations $6C-18D-6C-2D=1$ $D=-\frac{1}{20}$ $2C-6(-\frac{1}{20})=0$ $C=-\frac{3}{20}$ Therefore $y_{p}=-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$ $y=y_{c}+y_{p}$ $y=c_{1}e^{-2x}+c_{2}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$ The first condition is $y(0)=1$ $1=c_{1}+c_{2}-\frac{1}{4}-\frac{1}{20}$ $\frac{13}{10}=c_{1}+c_{2}$ (1) The second condition is $y'(0)=0$ $y'=-2c_{1}e^{-2x}+c_{2}e^{x}-\frac{1}{2}-\frac{6}{20}cos2x+\frac{2}{20}sin2x$ $0=-2c_{1}+c_{2}-\frac{1}{2}-\frac{6}{20}$ $\frac{2}{5}=-c_{1}+\frac{1}{2}c_{2}$ (2) Add equation (1) and (2) together $\frac{3}{2}c_{2}=\frac{13}{10}+\frac{4}{10}=\frac{17}{10}$ $c_{2}=\frac{17}{15}$ Substitute $c_{2}=\frac{17}{15}$ in equation (1) $c_{1}=\frac{13}{10}-\frac{17}{15}=\frac{39}{30}-\frac{34}{30}=\frac{5}{30}=\frac{1}{6}$ The solution of the initial problem is $y=\frac{1}{6}e^{-2x}+\frac{17}{15}e^{x}-\frac{1}{4}-\frac{x}{2}-\frac{3}{20}sin2x-\frac{1}{20}cos2x$
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