Calculus: Early Transcendentals 8th Edition

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

Chapter 9 - Section 9.1 - Modeling with Differential Equations - 9.1 Exercises - Page 590: 5

Answer

(a) This function is not a solution. (b) This function is not a solution. (c) This function is not a solution. (d) This function is a solution.

Work Step by Step

(a) $y = sin~x$ $y' = cos~x$ $y'' = -sin~x$ We can evaluate the differential equation: $y''+y = -sin~x+sin~x = 0 \neq sin~x$ This function is not a solution. (b) $y = cos~x$ $y' = -sin~x$ $y'' = -cos~x$ We can evaluate the differential equation: $y''+y = -cos~x+cos~x = 0 \neq sin~x$ This function is not a solution. (c) $y = \frac{1}{2}x~sin~x$ $y' = \frac{1}{2}~sin~x+ \frac{1}{2}x~cos~x$ $y'' = \frac{1}{2}~cos~x + \frac{1}{2}~cos~x-\frac{1}{2}x~sin~x$ $y'' = cos~x-\frac{1}{2}x~sin~x$ We can evaluate the differential equation: $y''+y = (cos~x-\frac{1}{2}x~sin~x)+(\frac{1}{2}x~sin~x) = cos~x \neq sin~x$ This function is not a solution. (d) $y = -\frac{1}{2}x~cos~x$ $y' = -\frac{1}{2}~cos~x + \frac{1}{2}x~sin~x$ $y'' = \frac{1}{2}~sin~x + \frac{1}{2}~sin~x+\frac{1}{2}x~cos~x$ $y'' = sin~x+\frac{1}{2}x~cos~x$ We can evaluate the differential equation: $y''+y = (sin~x+\frac{1}{2}x~cos~x)+(-\frac{1}{2}x~cos~x)= sin~x$ This function is a solution.
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.