How to solve a system of ODE's using laplace transform? System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions :  $$y_1(0)=2,\space y_1^{'}(0)=12$$
$$y_2(0)=6, \space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are: 
$$ s^2 \mathbf Y_1(s)-12s-2=16 \mathbf Y_2(s)$$
$$ s^2 \mathbf Y_2(s)-4s-6=16 \mathbf Y_1(s)$$
Then I put $\mathbf Y_2(s)$ in terms of $\mathbf Y_1(s)$
This gave me $$ \mathbf Y_2(s)=\frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?

If you do partial fractions this is what you should get
$$
{\bf Y}_2(x) = \frac{9}{2}\frac{1}{s - 4} + \frac{7}{2}\frac{1}{s + 4} - 2\frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = \frac{9}{2}e^{4t} + \frac{7}{2}e^{-4t} - \frac{1}{2}\sin 4t + 4\cos 4t
$$

EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
\begin{eqnarray}
\frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& \frac{A}{s - 4} + \frac{B}{s + 4} + \frac{Cs + D}{s^2 + 16}\\
&=& \frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \\
&=& \frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \\
&& + \frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \\
&=& \frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \\
&& + \frac{64A - 64B - 16D}{s^4 - 256}
\end{eqnarray}
And from here you conclude
\begin{eqnarray}
A + B + C &=& 4 \\
4A - 4B + D &=& 6\\
16A + 16B - 16C &=& 192 \\
64A - 64B - 16D &=& 32
\end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
