Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - Chapter Review Exercises - Page 525: 2



Work Step by Step

Since $y=x-2+e^{cx}$, then we have $y'=1+ce^{cx}$. Now, by substitution, we get $$2y'+y=2+2ce^{cx}+x-2+e^{cx}=x$$ which leads to $$(2c+1)e^{cx} =0$$ that is, $2c+1=0$ and hence $c=-\frac{1}{2}$.
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.