Showing integral curves are logarithmic spirals $\dot x:= \frac{\partial x}{\partial t}$, $\dot y:= \frac{\partial y}{\partial t}$
$$\dot x=-x+2y\\\dot y =-2x-y$$
I want to show that the integral curves of these ODEs are logarithmic spirals. We defined logarithmic spirals as $c: \mathbb R \to \mathbb R^2, t \mapsto (e^{at}\cos(bt),e^{at}\sin(bt)), a,b\neq 0$.
I tried transforming into polar coordinates $(r,\phi$) but I come to a reasonable solution 
 A: $$\begin{cases}\frac{dx}{dt}=-x+2y\\\frac{dy}{dt} =-2x-y\end{cases}$$
$$\frac{dy}{dx}=\frac{2x+y}{x-2y}$$
In polar coordinates :
$$\begin{cases}
x=r\cos(\phi)\quad;\quad dx=dr\cos(\phi)-r\sin(\phi)d\phi\\
y=r\sin(\phi)\quad;\quad dy=dr\sin(\phi)+r\cos(\phi)d\phi
\end{cases}$$
$$\frac{dr\sin(\phi)+r\cos(\phi)d\phi}{dr\cos(\phi)-r\sin(\phi)d\phi}=\frac{2r\cos(\phi)+r\sin(\phi)}{r\cos(\phi)-2r\sin(\phi)}$$
$$\frac{\frac{dr}{rd\phi}\sin(\phi)+\cos(\phi)}{\frac{dr}{rd\phi}\cos(\phi)-\sin(\phi)}=\frac{2\cos(\phi)+\sin(\phi)}{\cos(\phi)-2\sin(\phi)}$$
Let $\frac{dr}{rd\phi}=X$ and $\tan(\phi)=T$
$$\frac{XT+1}{X-T}=\frac{2+T}{1-2T} \quad\implies\quad X=\frac12$$
$$\frac{dr}{rd\phi}=\frac12$$
$$\ln|r|=\frac12\phi+c$$
$$r=Ce^{\phi/2}$$
This is the polar equation of a logarithmic spiral : http://mathworld.wolfram.com/LogarithmicSpiral.html
A: Let $z:=x+iy$. The system becomes
$$\dot z=(-1-2i)z,$$ which integrates as
$$z=z_0\,e^{(-1-2i)t}=|z_0|e^{(-1-2i)t+i\angle z_0}.$$
We have $a=-1,b=-2$ and there is an extra "phase" constant.

In polar coordinates,
$$\dot\rho\cos\theta-\rho\sin\theta\,\dot\theta=-\rho\cos\theta+2\rho\sin\theta,
\\\dot\rho\sin\theta+\rho\cos\theta\,\dot\theta=-2\rho\cos\theta-\rho\sin\theta.$$
Then by elimination,
$$\dot \rho=-\rho,\\\rho\,\dot\theta=-2\rho$$ which are trivial to integrate.
