Consider the coupled system $$\dot{x}=-x+y, \,\,\, \dot{y}=-x-y.$$ Working with polar coordinates $r$ and $\theta$ such that $$x=r\cos\theta, \,\,\, y=r\sin\theta,$$ Obtain uncoupled differential equations for $r$ and $\theta$.
I did some research online and found out how to convert a differential equation to polar coordinates. I used the relationship $$r^2 = x^2+y^2$$ Then I differentiated this equation, and plugged in our $\dot{x}$ and $\dot{y}$ values given.
I got $\dot{r}=-r.$ Which I'm nearly certain is correct. Then to find $\dot{\theta}$ I used $\tan \theta = \dfrac{r\sin\theta}{r\cos\theta} =\dfrac{y}{x},$ differentiate it and I found $\dot{\theta} = -1$.
Solve the two differential equations from (i) to obtain $r(t)$ and $\theta(t)$
This seems ok, I just integrated my $\dot{r}$ and my $\dot{\theta}$ equations. I got $$r(t) = -\dfrac{r^2}{2}\,\,\, \text{and} \,\,\, \theta(t)= -\theta$$
Using the results from (ii) sketch the trajectory and identify the type and stability of the equilibrium point $q=(0,0)$.
This is the part that I'm writing the question about. I just want make sure I can understand this right. Our $\theta$ is negative does that mean it will be spiralling anti-clockwise. Also $r$ is negative so does that mean it will be decaying towards the equilibrium? If that's the case wouldn't we have a stable attractive node around the equilibrium?
Thanks very much.
Your solutions of the differential equations are both wrong. The independent variable is $t$. $\dot r=-r$ has the well-known solution $r(t)=r_0·e^{-t}$ and $\dot θ=-1$ trivially $θ(t)=θ_0-t$.
Yes, you are right that the spirals are clock-wise. As the radius gets smaller in time, the image of the curve is an anti-clockwise spiral. This more intuitive interpretation can belongs to the time-inversed problem.