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: 18


$$ y =\frac{-t^2}{2 +ct^2}.$$

Work Step by Step

We have $$ t^3 y'+4y^2=0\Longrightarrow y^{-2}dy=-4t^{3}dt\\ \Longrightarrow\int y^{-2}dy=\int -4t^{3}dt+c\\ \Longrightarrow -y^{-1}=2t^{-2}+c \Longrightarrow y =\frac{-1}{2t^{-2}+c}.$$ Hence, the general solution is $$ y =\frac{-t^2}{2 +ct^2}.$$
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.