Phase portrait of a $2 \times 2$ system of linear, autonomous differential eqns. with a zero eigenvalue Let $\mathbf{Y} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$ and $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a, b, c, d \in \mathbb{R}$. Now, consider the system
$$
\frac{\operatorname d \mathbf{Y}}{\operatorname d t} = \mathbf{A Y}
$$
and let $\lambda_1, \lambda_2$ be its eigenvalues with $\lambda_1 = 0$ and $\lambda_2 \ne 0$, and $\mathbf{V}_1, \mathbf{V}_2$ the eigenvectors associated with them respectively. Then it can be seen that the general solution is
$$
\begin{equation}\tag{1}
\mathbf{Y} = k_1 \mathbf{V}_1 + k_2 e^{\lambda_2 t} \mathbf{V}_2\text{.}
\end{equation}
$$
Now, I understand that there is a line of equilibria along the $\mathbf{V}_1$ eigenvector. However, I do not understand why the rest of the phase portrait looks like this ($\lambda_2 > 0$ in this case): 
That is, why are the rest of the solutions lines parallel to $\mathbf{V}_2$ in the phase plane? Can anyone please shed some light as to how I might infer this from looking at $\text{(1)}$.

If you let $t' = k_2 e^{\lambda_2 t}$, then $(1)$ becomes
$$Y = k_1 V_1 + t' V_2.$$ This is the equation for a line through $k_1V_1$ with direction $V_2$ ie lines parallel to $V_2$.
