Taking Laplace transform of both sides and appying initial conditions of `y(0) = 1` and `y^(\ ')(0) = 0` gives:

`{(s^2Y-sy(0)-y^(\ ')(0)}+2{sY-y(0)}+5Y=0`

`(s^2Y-s)+2(sY-1)+5Y=0`

`(s^2+2s+5)Y=s+2`

Solving for Y and completing the square on the denominator gives:

` {: (Y,=(s+2)/(s^2+2s+5)),(,=(s+2)/((s^2+2s+1)+4)),(,=(s+2)/((s+1)^2+4)),(,=(s+1)/((s+1)^2+4)+1/2 2/((s+1)^2+4)):}`

Now, finding the inverse Laplace Transform gives us the solution for y as a function of t:

`y=e^(-t)cos\ 2t+1/2e^(-t) sin\ 2t`