ty’ + 2y = sin(t), t > 0
y' + (2/t)y = sin(t)t^(-1)
p(t) = 2/t
μ = e^( ∫2/t dt) = e^(2ln(t)) = t^2
Multiplying both sides by μ
y’ t^2 + 2ty = sin(t)t
(yt^2)’ = sin(t)t
Let u = t, dv = sin(t)dt
then du = dt, v = -cos(t)
yt^2 = - tcos(t) + ∫cos(t)dt
yt^2 = - tcos(t) + sin(t) + C
y = (- tcos(t) + sin(t) + C) / t^2, t > 0
