y’ + [(t+1)/t]y = 1
U = e^[\Int(t+1/t)]dt = te^t
te^t*y’ + (t+1)e^t*y = te^t
te^t*y = \Int(te^t)dt
u = t du = dt
dv = e^tdt v = e^t
te^t*y = te^t - \Int(e^t)dt
te^t*y = te^t - e^t + C
y = 1- 1/t + C/te^t
When t =ln2 y = 1- 1/ln2 + C/2ln2
C = 2
y = 1- 1/t(1+ 2/e^t)
When t > 0 and t —> infinite, y —> 1
