Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.3 Exercises: 26

Answer

$$y=\frac{1}{4}\sqrt{25-9x^2}$$

Work Step by Step

$y'=-\frac{9x}{16y}$ $16yy'=-9x$ $\int 16ydy=-\int 9xdx$ $8y^2=-9\frac{x^2}{2}+C$ $y^2=C-\frac{9x^2}{16}$ $y=\sqrt{C-\frac{9x^2}{16}}$ Now we need our equation to pass through the point $(1,1)$ so we apply it as an initial condition. $1=\sqrt{C-\frac{9}{16}}$ $1=C-\frac{9}{16}$ $C=\frac{25}{16}$ So $y=\sqrt{\frac{25}{16}-\frac{9x^2}{16}}$ $$y=\frac{1}{4}\sqrt{25-9x^2}$$
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.