Find the solution of the given initial value problem y'+2/ty=cost/t^2

asked Jan 12, 2018 213k views

2 votes

7.5k points

Please log in or register to add a comment.

5 votes

$y'+\frac2ty=(\cos t)/(t^2)$

$t^2y'+2ty=\cos t$

$(t^2y)'=\cos t$

$t^2y=\sin t+C$

$y=(\sin t+C)/(t^2)$

Sepero answered Jan 18, 2018

8.1k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.2m questions

11.9m answers