Cant solve this differential equation using laplace transforms $$\left\{\begin{array}{ccc}
 y''(t) &=& x(t) - 2y(t)\\
 x''(t) &=& - 2x(t) + y(t)
\end{array}\right.$$
$y(0)=1,\;x(0)=1,\; y'(0) = \sqrt3,\; x'(0) = -\sqrt3$.
I am trying to solve this differential system using laplace transforms but the terms make it very hard to calculate it. I got 
$$
x = \mathcal{L}^{-1} \left\{\frac{s^3 - \sqrt{3} s^2 + 3s - \sqrt{3}}{(s^2+2)^2 - 1} \right\}
$$ 
But that looks insanely difficult to find the laplace inverse. Is there anyway to make this simpler to do by hand?

Taking the LT of both equations and initial values yields
$$(s^2+2) Y(s)-X(s) = s+\sqrt{3}$$
$$(s^2+2) X(s)-Y(s) = s-\sqrt{3}$$
Solution yields
$$X(s) = \frac{s}{s^2+1} - \frac{\sqrt{3}}{s^2+3} $$
$$Y(s) = \frac{s}{s^2+1} + \frac{\sqrt{3}}{s^2+3} $$
Inversion of these, individually, yields
$$x(t) = \cos{t} - \sin{\sqrt{3} t}$$
$$y(t) = \cos{t} + \sin{\sqrt{3} t}$$
Plugging this solution into the original equation yields agreement.
ADDENDUM
The algebra involved in the solution is a little tricky.  To get $Y$, multiply both sides of the first equation by $s^2+2$ and add to the second equation to get
$$\left [ (s^2+2)^2-1\right ] Y(s) = (s^2+2) (s+\sqrt{3}) + s-\sqrt{3} = s(s^2+3) + \sqrt{3} (s^2+1)$$
Note that $(s^2+2)^2-1 = (s^2+1) (s^2+3)$.  Dividing through yields the solution.
To get $X$, multiply the second equation through by $s^2+2$ and add to the first equation.  The manipulations are virtually identical.
