The given general solution for some linear system is $ x= c_{2} \mathbf{ x^{(1)} }(t) + c_{2} \mathbf{ x^{(2)} }(t) = c_{1}\ \left(\begin{array}{l} 1\\ \sqrt{2} \end{array}\right)\ e^{-t}+ c_{2} \left(\begin{array}{l} -\sqrt{2}\\ 1 \end{array}\right)\ e^{-4t} \qquad (21) $.
Graphs of the solution (21) for several values of $c_{1}$ and $c_{2}$ are shown in Figure 7. $5.4a$.
The solution $x^{(1)}(t)$ approaches the origin along the line $x_{2}=\sqrt{2}x_{1}$, and the solution $x^{(2)}(t)$ approaches the origin along the line $x_{1}=-\sqrt{2}x_{2}$....
(1) Please explain this last sentence? I know $\color{brown}{ x_2 = \sqrt{2}x_{1} \iff } \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix}$ and $ \color{green} { x_{1}=-\sqrt{2}x_{2} } \iff \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} -\sqrt{2} \\ 1 \end{bmatrix}$
As $ t\rightarrow\infty$, the solution $x^{(2)}(t)$ is negligible compared to $x^{(1)}(t)$ .
(2) Why 'negligible'? I know when $k > 0$, $\lim_{t \to \infty} e^{-kt} = 0 $
Thus, unless $c_{1}=0$, the solution (21) approaches the origin tangent to the line $x_{2}=\sqrt{2}x_{1}$.
(3) Please explain this last sentence? I know if $c_1 =0 $ then any solution must be $ c_{2} \mathbf{ x^{(2)} }(t)$
... If the eigenvalues were positive rather than negative, then the trajectories would be similar but traversed in the outward direction.
(4) Please explain this last sentence?
The first equation tells us (implicitly) $$\mathbf{x}^{(1)}(t)= \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}e^{-t}, \quad \mathbf{x}^{(2)}(t)= \begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}e^{-4t}.$$
This means that, starting with $t=0$ and using $e^0 = 1$, we have initial values
$$\mathbf{x}^{(1)}(0)= \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}, \quad \mathbf{x}^{(2)}(0)= \begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}.$$
As $t$ increases from $0$ towards $\infty$, the factors $e^{-k}$ and $e^{-4k}$ decrease from $1$ to $0$, i.e. the solutions $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ run along the red/green line. In view of this, the black arrows are misleadingly positioned.
This argument of Boyce is a bit handwavy here, and my explanation is going to be similarly handwavy - I hope that this makes it more understandable. The idea is that $e^{-4t}$ converges faster to $0$ than $e^{-t}$ does. I would not call it negligible, but the two solutions converge to zero at different rates. One way to see this is to write $c_1 \mathbf{x}^{(1)}(t)+ c_2 \mathbf{x}^{(2)}(t)$ as
$$e^{-t}\left(c_1\begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix} + c_2\begin{pmatrix} -\sqrt{2}\\ 1\end{pmatrix}e^{-3t}\right) .$$
As $t$ increases, the second summand in the brackets goes to $0$, while the first one remains.
Because of this difference in convergence speed, a linear combination of the two solutions gets very close to $\mathbf{x}^{(1)}(t)$, as the part with $\mathbf{x}^{(2)}(t)$ 'converges away'. You can see this behavior in the blue arrows, which are the general solutions of the system. All of them approach the origin along the direction of $\pm \begin{pmatrix} 1 \\ \sqrt{2}\end{pmatrix}$ eventually - the part of $\mathbf{x}^{(2)}(t)$ dimishes away. I am aware that this explanation is similarly handwavy as the original one.
This is lacking context. How are the solutions related to eigenvalues (of which matrix)?