Toronto Math Forum

A typed solution may be more helpful
2(a):
\begin{equation*} \textbf{x}'=\begin{pmatrix}\hphantom{-}0 & 1\\\hphantom{-}2 &1 \end{pmatrix}\textbf{x}\ . \end{equation*}

find eigenvalues

\begin{equation*} r^2 - trace(A) + (ad- bc) =  r^2 -r - 2 = 0\implies r_1= 2, r_2=-1\end{equation*}

then, find eigenvectors, when r = 2

\begin{equation*} \begin{pmatrix} 0 - 2 & \hphantom{-}1\\  \hphantom{-}2 &1 -2\end{pmatrix}\begin{pmatrix}\mathbf{\xi}{^1}{_1}\\\mathbf{\xi}{^1}{_2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

then

\begin{equation*}\mathbf{\xi}^1 =\begin{pmatrix}1\\2\end{pmatrix}\end{equation*}

when r = -1

\begin{equation*} \begin{pmatrix} 0  + 1 & \hphantom{-}1\\  \hphantom{-}2 &1 +1\end{pmatrix}\begin{pmatrix}\mathbf{\xi}{^2}{_1}\\\mathbf{\xi}{^2}{_2}\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{equation*}

then

\begin{equation*}\mathbf{\xi}^2=\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}

Therefore

\begin{equation*}\mathbf{x}(t)= C_1e^{2t}\begin{pmatrix}1\\2\end{pmatrix}+ C_2e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}

Real eigenvalues with distinct signs, the type of origin is a saddle point.

2(b):

\begin{equation*} \mathbf{x}(0)=C_1+ C_2=2\\\mathbf{y}(0)=2C_1- C_2=1 \end{equation*}
Easy Calculation:
\begin{equation*} C_1 = 1, C_2 = 1 \end{equation*}
final answer:
\begin{equation*}\mathbf{x}(t)= e^{2t}\begin{pmatrix}1\\2\end{pmatrix}+ e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}\end{equation*}

I think two real Eigenvalue with same sign should give us a stable node. The direction in this case would be towards origin, since both x1 and x2 approach 0 as t tends to infinity.