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

You can tackle it, but it's not necessarily pretty, and as you've noticed - you can solve it perfectly fine in Cartesian coordinates so that's probably how you're expected to approach it.
However, if you're keen, then you can start like this (I'm using dots for time derivatives by convention and because $\theta'$ looks weird to me):
First, use $\dot{x} = \dot{r} \frac{\partial x}{\partial r} + \dot{\theta} \frac{\partial x}{\partial \theta} = \dot{r} \cos \theta - r \dot{\theta} \sin \theta$ and similarly $\dot{y} = \dot{r} \sin \theta + r \dot{\theta} \cos \theta$. Substituting those into the DE gives:
$$\begin{eqnarray}
\dot{r} \cos \theta - r \dot{\theta} \sin \theta & = & 2x - y & = & 2r \cos \theta - r \sin \theta \\
\dot{r} \sin \theta + r \dot{\theta} \cos \theta & = & 5x - 2y & = & 5r \cos \theta - 2r \sin \theta
\end{eqnarray}$$
Then you solve these as simultaneous equations for $\dot{r}$ and $\dot{\theta}$. I'll start with $\dot{\theta}$, dividing the first equation by $\cos \theta$ and the second by $\sin \theta$ and subtracting the first from the second (and being happy that I can cancel an $r$ term all around):
$$\begin{eqnarray}
\dot{r} - r \dot{\theta}\frac{\sin \theta}{\cos \theta} & = & 2r - r \frac{\sin \theta}{\cos \theta} \\
\dot{r} + r \dot{\theta}\frac{\cos \theta}{\sin \theta} & = & 5r \frac{\cos \theta}{\sin \theta} - 2r \\
r \dot{\theta} \left(\cot \theta - \tan \theta\right) & = & 5r \cot \theta - r \tan \theta - 4r \\
\dot{\theta}\frac{\cot \theta - \tan \theta}{5 \cot \theta -  \tan \theta - 4} & = & 1 \\
\int \frac{\cot \theta - \tan \theta}{5 \cot \theta -  \tan \theta - 4} \ d \theta & = & \int dt \\
t & = & \int \frac{\cot \theta - \tan \theta}{5 \cot \theta -  \tan \theta - 4} \ d \theta
\end{eqnarray}$$
and then you just have to calculate that integral, invert it to get $theta$ in terms of $t$, and then substitute that back into the DE somewhere to solve for $r$. And then, if you really want, you can then try to transform those back into $x$ and $y$.
Is it going to be nice? Uh, probably not. And there are also some singularities that you'll want to be careful about. Is it a good exercise? It probably depends a lot on how much you enjoy this kind of thing.
