Calculus (3rd Edition)

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

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 17


$$ y =\ln(4t^5+c)$$

Work Step by Step

We have $$\frac{dy}{dt}-20t^4e^{-y}=0\Longrightarrow e^{y}dy=20t^4\\ \Longrightarrow\int e^{y}dy=\int 20t^4dt+c\\ \Longrightarrow e^{y} =4t^5+c \Longrightarrow y =\ln(4t^5+c).$$ Hence, the general solution is $$ y =\ln(4t^5+c)$$
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.