How bout this one?

First, the matrix A is looked at:

$A = \left[ \begin{matrix} 0&-t\\t&0 \end{matrix}\right]$

The eigen values and their corresponding eigen vectors are:

$\lambda_1 = it \ ; \ \mathrm{v}_1 = \frac{1}{\sqrt{2}} \left[ \begin{matrix} i\\1\end{matrix}\right]$ $\lambda_2 = -it \ ; \ \mathrm{v}_2 = \frac{1}{\sqrt{2}} \left[ \begin{matrix} -i\\1\end{matrix}\right]$

We define:

$V = \left[ \begin{matrix}\mathrm{v}_1 & \mathrm{v}_2 \end{matrix}\right]$

From the eigen value decomposition of a matrix, we know that:

$A = VDV^{-1}$

Now, the matrix exponential:

$\mathrm{e}^{A} = V \mathrm{e}^{D} V^{-1}$ $\implies \mathrm{e}^{A}= V\left[ \begin{matrix} \mathrm{e}^{it} &0\\0&\mathrm{e}^{-it} \end{matrix}\right] V^{-1}$

We can see that the matrix $V$ is an orthogonal matrix, so its inverse is simply the complex conjugate transpose of $V$ . Using all this information, plugging into the above expression, crunching out the successive matrix multiplications and simplifying leads to:

$\mathrm{e}^{A} = \left[ \begin{matrix} \cos{t}&-\sin{t}\\ \sin{t}&\cos{t} \end{matrix}\right]$

Finally:

$\mathrm{e}^{A} \left[ \begin{matrix} \frac{1}{2}\\ \frac{\sqrt{3}}{2}\end{matrix}\right] =\frac{1}{2}\left[ \begin{matrix} \cos{t}-\sqrt{3}\sin{t} \\ \sin{t}+\sqrt{3}\cos{t}\end{matrix}\right] = \left[ \begin{matrix} -\sin\left(t - \frac{\pi}{6}\right) \\ \cos\left(t - \frac{\pi}{6}\right)\end{matrix}\right]$

Therefore:

$x = -\int \sin\left(t - \frac{\pi}{6}\right) \ dt$ $y = \int \cos\left(t - \frac{\pi}{6}\right) \ dt$

The result is the parametric equation of a circle.

I will show you how I obtained this odd-looking equation. Assume a parametrization of some curve that has the following properties that are consistent with a circle:

1) constant speed with respect to the parameter

2) constant magnitude of acceleration with respect to the parameter

Differentiate $(x'(t))^2+(y'(t))^2=c$ to obtain $x''x'+y''y' = 0.$

The solution to this equation is $x''=-f(t)y', y''=f(t)x'.$

In order for the acceleration to remain constant in magnitude,

$d=(-f(t)y')^2+(f(t)x')^2 = f(t)^2c,$

where $d$ is a constant.

Therefore $f(t)=\pm\sqrt{\frac{d}{c}}.$ Hence, $f(t)$ must be constant in this case.

Let $k=\pm\sqrt{\frac{d}{c}}.$

This leads to the following system of equations:

$\begin{bmatrix} x''(t)\\ y''(t) \end{bmatrix} =\begin{bmatrix} 0 & -k \\ k & 0 \end{bmatrix} \begin{bmatrix} x'(t)\\ y'(t) \end{bmatrix}.$

Assume $k=1$ and move on.

I will write this equation as $(X')'=AX'\Rightarrow (X')'-AX'=0\Rightarrow e^{-\int A}(X')'-e^{-\int A}AX'=0.$

$\Rightarrow (e^{-\int A}X')'=0 \Rightarrow e^{-\int A}X' = C \Rightarrow X' = e^{\int A}C = e^{At}C.$

Integrating both sides of the equation leads to the solution:

$X = \int e^{\begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}} Cdt,$ where C, of course, is a vector of constants.