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: 47


$$ t=\pm \sqrt{ \pi+4} $$

Work Step by Step

Since $$\sin y+y=t^{2}+C$$ is a solution and $y(2)=0$, then $$\sin (0) +0 =4+C\ \ \Rightarrow C=-4 $$ Hence the particular solution is $$\sin y+y=t^{2}+4$$ We find the $t$ that satisfies $ y( t)=\pi $, \begin{align*} \sin \pi+\pi&=t^{2}-4\\ \pi+4&=t^2 \end{align*} Then $$ t=\pm \sqrt{ \pi+4}$$
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.