Calculus: Early Transcendentals 8th Edition

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

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 925: 75

Answer

The function $u=e^{-\alpha^2 k^2 t} \sin (kx)$ is a solution of the heat conduction equation $u_t=\alpha^2 u_{xx}$.

Work Step by Step

Here, we have $u_x=k e^{-\alpha^2 k^2 t} \cos (kx)$ We need to take the second derivative of $u$ with respect to $x$. $u_{xx}=-k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$ ...(a) and $u_{t}=-\alpha^2 k^2 \times e^{-\alpha^2 k^2 t} \sin (kx)$...(b) From the above two equations (a) and (b), we have $u_t=\alpha^2 u_{xx}$ This is the required relationship for the heat conduction.
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.