Expressing an oscillator as a series of ODEs Consider an oscillator satisfying the initial value problem $u''+w^2u=0$, where $u(0)=u_0$, $u'(0)=v_0$. Let $x_1 = u$, $x_2=u'$, and transform the equations given into the form $x' = Ax, x(0)$.  Then using $$\exp(At) = I + \sum_{n=1}^\infty {A^nt^n\over n!}$$ show that $$\exp(At)= I\cos(wt) + A({\sin(wt)\over w})$$

I've gotten as far as substituting $x_1$ and $x_2$ for $u$, but I am not sure what to do next.  I have $x_2'+ w^2x_1=0$.  Any help you can give will, as always, be greatly appreciated.
 A: We are given the second order system:
$$\tag 1 u''+ \omega^2~u=0$$
We have:


*

*$x_1 = u$

*$x_1' = u' = x_2$

*$x_2' = u'' = -\omega^2 u = -\omega^2 x_1$


We can now write the system in matrix form as $x'= Ax$, which yields:
$$x'(t) = \begin{bmatrix}x_1' \\ x_2'\end{bmatrix} = Ax = \begin{bmatrix} 0& 1 \\ -\omega^2 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$
If we solve for the eigenvalues of $|A-\lambda I| = 0$, we have:
$$\lambda_{1,2} = \pm i ~ \omega$$
These are complex conjugate eigenvalues with zero real part, so what does that tell us about the solution? Here is a phase portrait for $\omega = 1$:

Can you continue it from here?
If we write the matrix exponential, we have:
$$e^{At} = \begin{bmatrix} \cos(\omega t) & \dfrac{\sin(\omega t)}{\omega} \\ -\omega \sin(\omega t) & \cos(\omega t)\end{bmatrix}$$
Now, look at this form and the hint in your problem and what do you notice? However, I left the details for you to fill in as it is very important for you to understand!
A: Of course, given the second order system
$u'' + w^2 u = 0, \tag{1}$
the transformation
$x_1 = u, \tag{2}$
$x_2 = x_1' = u'  \tag{3}$
yields, upon setting
$x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \tag{4}$
and
$A = \begin{bmatrix} 0 & 1 \\ -w^2 & 0 \end{bmatrix}, \tag{5}$
the equation
$x' = Ax; \tag{6}$
this covers the first part of our OP Heath Huffman's question.  As with any constant-coefficient linear system, the solution may be written
$x(t) = e^{A(t - t_0)}x(t_0), \tag{7}$
i.e.
$X(t) = e^{A(t - t_0)} \tag{8}$
is a fundamental matrix solution of (6) with $X(t_0) = I$.
A quick and cute way to get at the second part of the problem is to define the matrix
$J = w^{-1}A = \begin{bmatrix} 0 & w^{-1} \\ -w & 0 \end{bmatrix} \tag{9}$
and observe that
$J^2 = w^{-2}A^2 = w^{-2}\begin{bmatrix} -w^2 & 0 \\ 0 & -w^2 \end{bmatrix} = -I. \tag{10}$
From (10) it follows (very nearly) immediately that, for any real $\theta$,
$e^{\theta J} = (\cos \theta) I + (\sin \theta) J, \tag{11}$
just as we have 
$e^{i \theta} = \cos \theta + i \sin \theta,  \tag{12}$
where $i^2 =-1$ is the ordinary imaginary unit.  (11) relies on two facts:  i.)  $J$ is a bounded linear map, so the series for $e^{\theta J}$,
$e^{\theta J} = \sum_0^{\infty} (\theta J)^n / {n!}, \tag {13}$
converges absolutely and uniformly on any bounded interval $[-M, M]$, where $M > 0$; indeed, the series (13) is majorized, term by term, by that for $e^{\vert \theta \vert \Vert J \Vert}$, where $\Vert J \Vert$ is a suitably chosen operator norm for the $2 \times 2$ matrix $J$:
$e^{\vert \theta \vert \Vert J \Vert} = \sum_0^{\infty} \vert \theta^n \vert \Vert J \Vert^n / n!; \tag{14}$
the requisite convergence of (13) then follows from (14); and (ii.) the cyclic properties of powers of $J$ and $i$ are identical; indeed we have
$J^2 = -I, \, J^3 = -J, \, J^4 = -J^2 = I, \, J^5 = J, \, J^6 = J^2 = -I, \, \text{etc.}, \tag{15}$
exactly as we also have
$i^2 = -1, \, i^3 = -i, \, i^4 = -i^2 = 1, \, i^5 = i, \, i^6 = i^2 = -1, \, \text{etc.} \tag{16}$
(15) and (16), together with the convergence of the series (13), establish that the terms of (13) may be grouped according to powers of $J$ exactly as those of 
$e^{i \theta} = \sum_0^{\infty} (i \theta)^n / n! \tag{17}$
are grouped according to powers if $i$, and that indeed the relevant signs correspond.  This establishes the veracity of (11).  Thus, setting $\theta = wt$, we obtain
$e^{wtJ} = (\cos wt) I + (\sin wt) J, \tag{18}$
and inserting (9) into (18) yields the desired formula
$e^{At}= e^{wt (w^{-1}A)} = e^{wt J} =\cos(wt)I + (\sin wt)J =(\cos wt)I + (w^{-1}\sin wt)A. \tag{19}$
QED.
I think the interesting thing about this approach is that it shows that, implicitly within the structure of the matrix $A$, is the real matrix $J = w^{-1}A$ which captures all the essential features of the imaginary unit $i$, including the cyclic periodicity properties (15), (16), which of course give rise to the essential Euler formulas (11), (12), and these features may be exploited without explicitly calculating the eigenvalues and eigenvectors of $A$.  In fact, the methodology emphasized here may be taken a step further, to the case where $A$ has a complex conjugate pair of eigenvalues $\lambda, \bar \lambda = \sigma \pm i \omega$, though to do so does in fact computing $\lambda, \bar \lambda$, a relatively simple undertaking.  It works like this:  it is well-known, by the Cayley-Hamilton theroem, that $A$ satisfies the equation
$A^2 - (\text{Tr} A) A + (\det A)I = 0, \tag{20}$
and if the eigenvalues are $\sigma \pm i \omega$, it is easy to see that
$\text{Tr} A = \lambda + \bar \lambda = 2 \sigma, \, \det A = \lambda \bar \lambda = \sigma^2 + \omega^2; \tag{21}$
(20) becomes
$A^2 - 2\sigma A + (\sigma^2 + \omega^2)I = 0, \tag{22}$
which may be written
$(A - \sigma I)^2 = -\omega^2 I \tag{23}$
or
$(\omega^{-1}(A - \sigma I))^2 = -I, \tag{24}$
so if we now set
$J = \omega^{-1}(A - \sigma I) \tag{25}$
we see that, as in the above,
$J^2 = -I.  \tag{26}$
From (26) it now follows, for exactly the same reasons as previously given, that (11) holds with $J$ as in (25).  Thus taking $\theta = \omega t$, in parallel with (18), (19) we may write
$e^{(A - \sigma I)t} = e^{\omega t(\omega^{-1}(A - \sigma I))} = e^{\omega tJ} = (\cos \omega t)I + (\sin  \omega t)J$
$= (\cos \omega t)I + (\omega^{-1}\sin \omega t)(A - \sigma I). \tag{27}$
Continuing in the same direction, we may exploit the fact that, since $[A, \sigma I] = 0$, i.e. $A$ commutes with $\sigma I$, to write
$e^{(A - \sigma I)t} = e^{-\sigma t I} e^{At} = (e^{-\sigma t}I) e^{At} = e^{-\sigma t} e^{At}, \tag{28}$
and if (28) is inserted into (27) we obtain
$e^{-\sigma t} e^{At} = (\cos \omega t)I + (\sin  \omega t)J = (\cos \omega t)I + (\omega^{-1}\sin \omega t)(A - \sigma I), \tag{29}$
whence
$e^{At} = e^{\sigma t}(\cos \omega t)I + e^{\sigma t}(\sin \omega t)J = e^{\sigma t}(\cos \omega t)I + e^{\sigma t}(\omega^{-1}\sin \omega t)(A - \sigma I), \tag{30}$
which generalizes (19) to the case $\lambda, \bar \lambda = \sigma \pm i \omega$.  The fact that  $[A, \sigma I] = 0$ implies (28) is proved by Yours Truly in my answer to this question, which was fortuitously and serendipitously posed just I was writing up the present answer; how convenient! ;-)!!!  In any event, these maneuvers illustrate how matrices such as $e^{At}$ exhibit a time evolution which exactly parallels that of $e^{\lambda t} = e^{(\sigma + i\omega)t} = e^{\sigma t}e^{i\omega t} = e^{\sigma t}(\cos \omega t + i\sin \omega t)$ provided we set $J = \omega^{-1}(A - \sigma I)$; the complex representation of solutions is implicitly inherent in the real matrix $A$, and the invarinant subspace or eigenspace corresponding $\lambda = \sigma + i \omega$ is manifested in the presence of $A$ and $J$ in these formulas, being as they are operators on $\Bbb R^2$, and not scalars.
