Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.1 Exercises - Page 403: 31


$4y^2= x^3$ passes through the point (4,4)

Work Step by Step

Use the general solution of the differential equation and plug in (4,4). $y^2= Cx^3$ $16= 64C$ $C=\frac{1}{4}$ Substitute C back into the general solution $ y^2= \frac{1}{4}x^3$ $4y^2= x^3$
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.