Question: ty' + 2y = sin(t), t>0 and determine how solutions behave as t->infinity
Answer: y' + (2/t)y = sin(t)/t
then p(t) = 2/t -> u(t) = e^(integral(p(t)dt) = t^2
multiply t^2 for both sides, we get: t^2y' + 2ty = sin(t)t
by using integration bt parts we have: y(t) = t^(-2)(sin(t) - cos(t)t + C)
Thus when t goes to infinity, the solution y(t) becomes 0.
(please see the pdf document attached below for details
)
