<s>[INST] 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?
 [/INST] 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.
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