Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 17 - Second-Order Differential Equations - 17.2 Exercises: 1

Answer

$y=c_{1}e^{-x}+c_{2}e^{3x}-\frac{7}{65}cos(2x)-\frac{4}{65}sin(2x)$

Work Step by Step

$y''-2y'-3y=cos(2x)$ Auxiliary equation: $r^{2}-2r-3=0$ $(r+1)(r-3)=0$ $r_{1}=-1$ $r_{2}=3$ $y_{c}=c_{1}e^{r_{1}x}+c_{2}e^{r_{2}x}$ $y_{c}=c_{1}e^{-1x}+c_{2}e^{3x}$ $y_{p}=Acos(2x)+Bsin(2x)$ $y_{p}'=-2Asin(2x)+2Bcos(2x)$ $y_{p}''=-4Acos(2x)-4Bsin(2x)$ Plug back into the original equation $(-4Acos(2x)-4Bsin(2x))-2(-2Asin(2x)+2Bcos(2x))-3(Acos(2x)+Bsin(2x))=cos(2x)$ $(-7A-4B)cos(2x)+(4A-7B)sin(2x)=cos(2x)$ thus $(-7A-4B)=1$ $(4A-7B)=0$ $A=-\frac{7}{65}$ $B=-\frac{4}{65}$ $y=y_{c}+y_{p}$ $y=c_{1}e^{-x}+c_{2}e^{3x}-\frac{7}{65}cos(2x)-\frac{4}{65}sin(2x)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.