<s>[INST] How to use polar coordinate in ODE? I don't understand how to use polar coordinate.
\begin{cases} \frac{dx(t)}{dt}=2x-y \\ \frac{dy(t)}{dt}=5x-2y  \\ \end{cases}
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \begin{pmatrix} 2 & -1\\ 5 & -2 \end{pmatrix} \left( \begin{array}{c} x \\ y \end{array} \right)$$
let $A = \begin{bmatrix} 2 & -1\\ 5 & -2 \end{bmatrix}$, we have $\det(A) = 1, \operatorname{tr}(A) = 0, P(\lambda) = \lambda^2+1 = 0 \Rightarrow \lambda_{1,2}=\pm i $
\begin{cases} 2x-y-ix = 0 \\ 5x-2y-iy = 0  \\ \end{cases}
x = 1, y = 0
$\vec v_1 = (x,y) = (x, x(2-i)) = x(1,(2-i)) = \binom{1}{2-i} $
$E_{\lambda1} = \binom{1}{0}+i \binom{0}{-1} = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix} = P $ which give us $P^{-1}= \begin{pmatrix} -1 & 0\\ -2 & 1 \end{pmatrix}$  and $J = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$
$X(t)= e^{0t} \begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
$Y(t) = \begin{pmatrix} 1 & 0\\ 2 & -1 \end{pmatrix}\begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
It's a center stable
So the solution is pretty simple but i want to try to use the polar coordinate and i can't do it
$r^2= x^2 + y^2$ and $x = r\cos(\theta)$, $y = r \sin(\theta)$
$2rr'= 2xx'+ 2yy' = 2x(2x-y) + 2y(5x-2y) = 4x^2 +8xy -4y^2 $
$\theta ' = \frac{y(2x-y)+x(5x-2y)}{x^2+y^2} = \frac{-y^2+5x^2}{x^2+y^2}$
maybe if the exercise doesnt ask polar coordinate, we don't do it ?
i've check this : Polar coordinates differential equation but i still can't do it
 [/INST] If you define, following the rows of $P^{-1}$, $u=x$ and $v=y-2x$, then the system becomes
$$
u'=x'=-v
\\
v'=y'-2x'=x=u
$$
which is the standard circle system, or rotation with constant angular speed of $-1$. That you get this structure you also found in the matrix $J$.
With this you get the constant radius
$$
R^2=u^2+v^2=x^2+(y-2x)^2.
$$
</s>