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: 12



Work Step by Step

$y''-6y'+25y=0$ which has characteristic equation $t^2-6t+25=0$ with solutions $t=3 \pm 4i$ General solution: $c_1e^{3x}cos(4x+c_2e^{3x}sin(4x)$ Plug in the values: $y(0)=2$ and $y'(0)=1$ giving $c_1=2$, $c_2=\frac{-5}{4}$ Hence, $y=2e^{3x}cos4x-\frac{-5}{4}e^{3x}sin4x$
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.