Change of variables and circular solutions Consider the equations:
$ \frac{dx}{dt} = y$  and  $ \frac{dy}{dt} = -x $
By transforming variables, obtain $ \frac{dr^2}{dt}=0 $ and $ \frac{d \theta}{dt} = -1$
I know that if I say let $r^2 =x^2 +y^2 \cdot \frac{dr^2}{dt} = tx \frac{2x}{dt} + 2y \frac{dy}{dt} =2xy =2xy-2xy = 0$, which would mean $r^2$ is neutrally stable, but I don't understand where this gets me, much less what a change of variables is. 
Moreover, what can one conclude about whether these are circular solutions, their direction, and their stability. I have not seen this mentioned before.
 A: Assuming you're using polar coordinates, a change of variables to the polar coordinates system corresponds to $x=r\cosθ,y=r\sin θ,r=x^2+y^2$ :
It is :
$$r^2 = x^2 + y^2 \Rightarrow 2rr' = 2xx' + 2yy' \Rightarrow rr' = xx' + yy' $$
Substituting $x',y'$ from your given system :
$$rr'= xy - yx = 0$$
To find the angle $θ$, we take : 
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule, we get :
$$\theta' = \dfrac{x y' - y x'}{r^2}$$
Substituting $x',y'$ as before, from your given system : 
$$θ = \frac{-x^2 -y^2}{r^2} =\frac{-(x^2 + y^2)}{r^2}=\frac{-r^2}{r^2}=-1$$
This is how you get the expressions.
Going over stability, it's easy to see that $O(0,0)$ is a stationary point of your given system and also the only one.
Since you have $rr' = 0$, you can conclude that $O(0,0)$ is clarified as a center for your system.
You can easily double check that by doing the common stability way :
$$J(x,y) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$
Which means that : 
$$J(0,0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$
and that : 
$$\det(J(0,0) - λI) = 0 \Leftrightarrow \cdots \Leftrightarrow  λ^2 + 1=0 \Leftrightarrow λ = \pm i$$
Since you have purely imaginary eigenvalues, $O(0,0)$ is a center for your system, thus we have double checked our finding.
Can you now make a conclusion about the direction and the circularity of the solutions ?
A: The best I can interpret the situation is as follows:
$dx/dt = y$ and $dy/dt = -x$ when solved (accompanied with initial and/or boundary conditions, which are absent from your problem statement), will give a solution of the form $x = x(t)$ and $y = y(t)$. When you plot this on x-y plane, you will get a curve, with $t$ being a parameter.
Every curve on x-y plane can be expressed by a polar equation as well, with $x = r \cos \theta$ and $y = r \sin \theta$. Now, in your case, $r$ and $\theta$ will also be functions of $t$. Use these transformations along with chain and product rules for differentiation to get $dr^2/dt = 0$ and $d \theta / dt = -1$.
The parameter $t$ in fact controls how fast a particle moves on that curve. You can have the exact same curve traced by two different particles at different speeds. A simple ODE in $x$ and $y$ will give you only the curve without any information about the speed of the particle.
A: with $$x'=y$$ and $$y'=-x$$ we get $$x''=-x$$ and you can solve this equation
