Cant solve this differential equation using laplace transforms

Question

$$\left\{\begin{array}{ccc} y''(t) &=& x(t) - 2y(t)\\ x''(t) &=& - 2x(t) + y(t) \end{array}\right.$$ $y(0)=1,\;x(0)=1,\; y'(0) = \sqrt3,\; x'(0) = -\sqrt3$.

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?

Answer 1

Note that $s^3-\sqrt3s^2+3s-\sqrt3=(s^2+1)(s-\sqrt3)+2s$ and that $(s^2+2)^2-1=(s^2+1)(s^2+3)$. Thus $$x=\mathcal L^{-1}\left(\frac{s-\sqrt3}{s^2+3}+\frac{s}{s^2+1}-\frac{s}{s^2+3}\right).$$ (I have used a decomposition into simple elements.) It is a simple inversion, yielding $$x(t)=\cos(t)-\sin(t\sqrt3).$$

Answer 2

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.