Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 9: First-Order Differential Equations - Section 9.1 - Solutions, Slope Fields, and Euler's Method - Exercises 9.1 - Page 531: 21


$y'=x+y$; $y(x_0)=y_0$

Work Step by Step

Here, $y'=-1+(1+x_0+y_0)e^{(x-x_0)}$ and $x+y=x+[-1-x+(1+x_0+y_0)e^{(x-x_0)}]$ This implies that $x+y=-1+(1+x_0+y_0)e^{(x-x_0)}$ Thus, this proves that $y'=x+y$ and $y(x_0)=y_0$
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.