Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises - Page 627: 37

Answer

$y=\frac{x^4}{4}+2x^2+4$

Work Step by Step

Find the initial condition: $y(0)=4+\int_0^02t\sqrt{y(t)}dt$ $y(0)=4+0$ $y(0)=4$ Construct a differential equation by differentiating the integral equation: $\frac{d}{dx}(y(x))=\frac{d}{dx}(4+\int_0^x2t\sqrt{y(t)}dt)$ $\frac{dy}{dx}=0+2x\sqrt{y(x)}$ $\frac{dy}{dx}=2x\sqrt{y}$ Solve for $y$ by separating variables: $\frac{1}{2\sqrt{y}}dy=xdx$ $\int \frac{1}{2\sqrt{y}}dy=\int xdx$ $\sqrt{y}=\frac{x^2}{2}+C$ Substitute the initial condition: $\sqrt{4}=\frac{0^2}{2}+C$ $2=0+C$ $C=2$ We have now: $\sqrt{y}=\frac{x^2}{2}+2$ $y=(\frac{x^2}{2}+2)^2$ $y=\frac{x^4}{4}+2x^2+4$ Thus, the solution is $y=\frac{x^4}{4}+2x^2+4$.
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.