Suppose \(\lambda_1 \lt \lambda_2 \lt 0\) and consider the diagonal system
\begin{equation*}
\begin{pmatrix}
x'(t) \\ y'(t)
\end{pmatrix}
=
\begin{pmatrix}
\lambda_1 & 0 \\
0 & \lambda_2
\end{pmatrix}
\begin{pmatrix}
x(t) \\ y(t)
\end{pmatrix}.
\end{equation*}
The general solution of this system is
\begin{equation*}
{\mathbf x}(t)
=
c_1 e^{\lambda_1 t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}
+
c_2 e^{\lambda_2 t} \begin{pmatrix} 0 \\ 1 \end{pmatrix},
\end{equation*}
but unlike the case of the saddle, all solutions tend towards the origin as \(t \to
\infty\text{.}\) To see how the solutions approach the origin, we will compute \(dy/dx\) for \(c_2 \neq 0\text{.}\) If
\begin{align*}
x(t) & = c_1 e^{\lambda_1 t}\\
y(t) & = c_2 e^{\lambda_2 t},
\end{align*}
then
\begin{equation*}
\frac{dy}{dx}
=
\frac{y'(t)}{x'(t)}
=
\frac{\lambda_2 c_2 e^{\lambda_2 t}}{\lambda_1c_1 e^{\lambda_1 t}}
=
\frac{\lambda_2 c_2}{\lambda_1c_1 } e^{(\lambda_2 - \lambda_1) t}.
\end{equation*}
Since
\(\lambda_2 - \lambda_1 \gt 0\text{,}\) the derivative,
\(dy/dx\text{,}\) must approach
\(\pm
\infty\text{,}\) provided
\(c_2 \neq 0\text{.}\) Therefore, the solutions tend towards the origin tangentially to the
\(y\)-axis (
Figure 5.3.6). We say that the equilibrium point for this system is a
sink.