Question

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

Answer 1

$y'+\dfrac2ty=\dfrac{\cos t}{t^2}$
$t^2y'+2ty=\cos t$
$(t^2y)'=\cos t$
$t^2y=\sin t+C$
$y=\dfrac{\sin t+C}{t^2}$