Prof, how about direction of improper node?
Don't hijack topics!OK, consider canonical form of $\mathbf{x}'=A\mathbf{x}$:
\begin{equation*}
\begin{pmatrix} x' \\ y'\end{pmatrix}= \begin{pmatrix}r & 1 \\ 0 &r\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}.
\end{equation*}
Then $y= Ce^{rt}$, $x= (Ct+C_1)e^{rt}$. Right? Depending on $r<0$ and $r>0$ you get one of two pictures (stable and unstable, respectively).
Now you need to learn if it is clock-wise or counter-clock-wise. Again the sign of the top-right element of the matrix defines it (clock-wise iff it is positive)
