Using a symmetric matrix to solve a second order DE Suppose I have the second order DE
$\frac{d^2}{dt^2}\underline{x}=-S\underline{x}(t)$
How could I use the knowledge that S is a symmetric matrix, $S=O^TDO$, to show that a general solution to the above is:
$\underline{x}(t) = [A\sin(\omega_it)+B\cos(\omega_it)]\underline{v_i}$
Any help would be greatly appreciated
 A: :) Hi d3V!
If your matrix is symmetric, it means, as you already mentioned, that you can diagonalize it, which is in such a case very useful.
For an $x(t) \in \mathbb{R}^n$, you have the following problem:
$$\dfrac{d^2x(t) }{dt^2}= -S x(t) \tag 1$$
And for such a problem in one dimension, we already know that a common solution is given by $x(t) = A'e^{ikx} + B'e^{-ikx}$, or equivalently $A \sin(kx) + B \cos(kx)$ for some $k \in \mathbb{R}$.
For our case, the diagonalisation is interesting, because otherwise, computing the exponential of matrices is really not funny. In fact, it holds:
$e^{-S x} = e^{-O^TDO x} = O^T e^{-D x} O$
Now, the space of solution to  $( 1)$, our initial problem, is given by:
$O^T e^{-D x} O = im(O^T e^{-D x})$
So you can come to the expression that you claimed, if you consider the vectors $v_i$ to be the columns of O^T, and the real values $\omega_i$ to be the diagonal elements of D.
With this elements, I hope you'll be able to finish your problem! :)
EDIT: the $\omega_i$ are the square root of the eigenvalues, not the eigenvalues themselves! sorry for the confusion ;)
