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`