Sketch $\dot x =x-y$, $\dot y= x+y$ on a phase portrait. Sketch $$\begin{cases}\dot x =x-y
\newline\dot y= x+y\end{cases}$$ on a phase portrait.
The original question requires me to change the variables to polar ones, but then asks me to draw a phase portrait of the problem using the prior form and then the new form, and I am currently stuck with drawing it, for we were never guided to use multiplication of matrices and vectors, but to actually solve the equation, which I just can't. I tried dozen of things and it seems like I always get stuck with the endless integration. I could use some help here.
 A: One analytical way to find the diagram Did has in his answer:
Writing $'$ for $d/dt$, then
$$xx' + yy' = x^2 + y^2$$
thus $\displaystyle \frac 12 (r^2)' = r^2$ and hence $r^2(t) = e^{2t}$ or $$r(t) = e^t$$
Meantime $\theta = \arctan(y/x)$ and 
$$\theta' = \frac{1}{1 + y^2/x^2}\left( \frac{y'}{x} - \frac{y}{x^2}  x'\right) = \frac{xy' - yx'}{x^2 + y^2} = \frac{x^2 + y^2}{x^2 + y^2} = 1$$
Therefore the general solution is
$$\begin{align} x(t) & = c_1 e^t \cos t + c_2 e^t \sin t \\
                y(t) & = c_1 e^t \sin t - c_2 e^t \cos t
\end{align}$$
Angular velocity constant, $\theta' = 1$; radial distance growing exponentially, $r(t) = e^t$.

The more sophisticated method is to find the eigenvectors and general solution of the system
$$\frac{d\ }{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{ccc}
1 & -1 \\
1 & 1 \\
\end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right)$$
The eigenvalues are $1 \pm i$, consistent with the general solution we have above.
A: Notice that there is only one equilibrium, $(0,0)$. If you find the Jacobian of this system, you will find that it is:$ \left( \begin{array}{ccc}
1 & -1 \\
1 & 1 \\
\end{array} \right) $. You will find that the real part of the (complex) eigenvalues are positive, which implies that the origin is unstable. And since it is complex, it will be a spiral. Proceed similarly for the polar one.
A: 
Look at the picture and the system and notice that:


*

*In the Yellow Area: $y>x \Rightarrow \dot{x}<0 \;\;\;$ and $\;\;\; y<-x \Rightarrow \dot{y}<0$.

*In the Green Area: $\; y>x \Rightarrow \dot{x}<0 \;\;\;$ and $\;\;\; y>-x \Rightarrow \dot{y}>0$.

*In the Blue Area: $\;\;\; y<x \Rightarrow \dot{x}>0 \;\;\;$ and $\;\;\; y>-x \Rightarrow \dot{y}>0$.

*In the Red Area: $\;\;\;\; y<x \Rightarrow \dot{x}>0 \;\;\;$ and $\;\;\; y<-x \Rightarrow \dot{y}<0$.

A: Here is a plot of the solutions, which WA calls streamplot. For an analytical solution explaining it, see the answer posted by @SimonS.

