Worked-out solution

$\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A =\matrix{3 & 0\\ 0 & -1}$

1. $\quad\left\{\begin{aligned} \frac{\dd x_1}{\dd t} &= 3x_1\\[0.2cm] \frac{\dd x_2}{\dd t} &= -x_2\end{aligned}\right.$
   
   
2. $\quad\left\{\begin{aligned} x_1(t) &= c_1\cdot e^{3t}\\[0.2cm] x_2(t) &= c_2\cdot e^{-t}\end{aligned}\right.$ with constants $c_1$ and $c_2$.
   
   
3. $\quad\vec{x}(0)=\cv{1\\0}$ means that $x_1(0)=1$ en $x_2(0)=0$.
   $\quad$Substitution in the general solution (b) gives $c_1=1$ and $c_2=0$, and thus $\vec{x}(t)=\cv{e^{3t}\\0}$.
   $\quad$Then: $\vec{x}(1)=\cv{e^{3}\\0},\quad \vec{x}(2)=\cv{e^{6}\\0}, \quad \vec{x}(4)=\cv{e^{12}\\0}, \quad \vec{x}(8)=\cv{e^{24}\\0}$.
   
   
4. $\quad\vec{x}(0)=\cv{0\\10}$ means that $x_1(0)=0$ and $x_2(0)=10$.
   $\quad$Substitution in the general solution (b) gives $c_1=0$ and $c_2=10$,
   $\quad$and thus $\vec{x}(t)=\cv{0\\10e^{-t}}=\cv{0\\ \frac{10}{e^{t}}}$.
   $\quad$Then: $\vec{x}(1)=\cv{0\\ \frac{10}{e}},\quad \vec{x}(2)=\cv{0\\ \frac{10}{e^2}}, \quad \vec{x}(4)=\cv{0\\ \frac{10}{e^4}}, \quad \vec{x}(8)=\cv{0\\ \frac{10}{e^8}}$.
   
   
5. $\quad\vec{x}(0)=\cv{0.1\\10}$ means that $x_1(0)=0.1$ and $x_2(0)=10$.
   $\quad$Substitution in the general solution (b) gives $c_1=0.1$ and $c_2=10$,
   $\quad$and thus $\vec{x}(t)=\cv{0.1e^{3t}\\10e^{-t}}=\cv{0.1e^{3t}\\ \frac{10}{e^{t}}}$.
   $\quad$Then: $\vec{x}(1)=\cv{0.1e^3\\ \frac{10}{e}},\quad \vec{x}(2)=\cv{0.1e^6\\ \frac{10}{e^2}}, \quad \vec{x}(4)=\cv{0.1e^{12}\\ \frac{10}{e^4}}, \quad \vec{x}(8)=\cv{0.1e^{24}\\ \frac{10}{e^8}}$.
   
   
6. $\quad\vec{x}(0)=\cv{\alpha\\ \beta}$ means that $x_1(0)=\alpha$ en $x_2(0)=\beta$.
   $\quad$Substitution in the general solution (b) gives: $c_1=\alpha$ en $c_2=\beta$.
   $\quad$Thus: $\vec{x}(t)=\cv{\alpha\cdot e^{3t}\\ \beta\cdot e^{-t}}=\cv{\alpha\cdot e^{3t}\\ \dfrac{\beta}{e^{t}}}$.
   $\quad$The above formula is the parameter representation of the general solution.
   $\quad$This solution satisfies the following equation: $x_1\cdot x_2^3=\alpha\beta^3$.
   $\quad$Thus, any solution satisfies $x_1\cdot x_2^3=c$ for some constant $c$.
   $\quad$This is illustrated in the phase portrait below.
   $\quad$

Worked-out solution

$\quad\frac{\dd}{\dd t}\vec{x}=A\, \vec{x}\quad$ with $A = \matrix{3 & -6\\0 & -1\\}$

1. $\quad$The characteristic polynomial of $A$ is because of its upper triangular shape
   $\quad$equal to $(\lambda-3)(\lambda+1)$
   $\quad$The eigenvalues of $A$ are zeros of the characteristic polynomial: $3$ and $-1$.
   
   $\quad$Let $\cv{v_1\\ v_2}$ be an eigenvector corresponding to the eigenvalue $3$:
   $\quad$$\matrix{3 & -6\\0 & -1\\}\cdot \cv{v_1\\ v_2}=3\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} 3v_1-6v_2 &= 3v_1 \\[0.2cm] -v_2 &= 3v_2\end{aligned}\right.$
   $\quad$Thus: $v_2=0$ and an eigenvector corresponding to the eigenvalue $3$ is $\cv{1\\0}$.
   
   $\quad$Let $\cv{v_1\\ v_2}$ be an eigenvector corresponding to the eigenvalue $-1$:
   $\quad$$\matrix{3 & -6\\0 & -1\\}\cdot \cv{v_1\\ v_2}=-\cv{v_1\\ v_2}$.
   
   $\quad$Then: $\left\{\begin{aligned} 3v_1-6v_2 &= -v_1 \\[0.2cm] -v_2 &= -v_2\end{aligned}\right.$
   
   $\quad$So: $v_2$ can be freely chosen and then $4v_1-6v_2=0$, i.e. $v_2=\tfrac{2}{3}v_2$.
   $\quad$An eigenvector corresponding to the eigenvalue $-1$ (with integral coefficients) is $\cv{3\\2}$.
   
   
2. $\quad$The general solution is $$\begin{aligned}\vec{x}(t)&=\alpha\cdot e^{3t}\cdot\cv{1\\0}+ \beta\cdot e^{-t}\cdot\cv{3\\2}\\ \\ &= \cv{\alpha\cdot e^{3t}+3\beta\cdot e^{-t}\\ 2\beta\cdot e^{-t}}\end{aligned}$$ $\quad$ for certain constants $\alpha$ and $\beta$.
   
   
3. $\quad\vec{x}(0)=\cv{1\\0}$ means $x_1(0)=1$ and $x_2(0)=0$.
   $\quad$Substitution in the general solution (b) gives $\alpha=1$ and $\beta=0$,
   $\quad$and thus $\vec{x}(t)=\cv{e^{3t}\\0}$.
   $\quad$Then: $\vec{x}(1)=\cv{e^{3}\\0},\quad \vec{x}(2)=\cv{e^{6}\\0}, \quad \vec{x}(4)=\cv{e^{12}\\0}, \quad \vec{x}(8)=\cv{e^{24}\\0}$.
   
   
4. $\quad$When $\vec{x}(0)=\cv{1\\ \frac{2}{3}}$ then $\vec{x}(0)=\frac{1}{3}\cv{3\\ 2}$.
   $\quad$Substitution in the general solution (b) gives $\alpha=0$ and $\beta=\frac{1}{3}$,
   $\quad$and thus $\vec{x}(t)=\frac{1}{3}e^{-t}\cdot\cv{3\\2}= \cv{\frac{1}{e^t}\\ \frac{2}{3e^t}}$.
   $\quad$Then: $\vec{x}(1)=\cv{\frac{1}{e}\\ \frac{2}{3e}},\quad \vec{x}(2)=\cv{\frac{1}{e^2}\\ \frac{2}{3e^2}}, \quad \vec{x}(4)=\cv{\frac{1}{e^4}\\ \frac{2}{3e^4}}, \quad \vec{x}(8)=\cv{\frac{1}{e^8}\\ \frac{2}{3e^8}}$.
   
   
5. $\quad\vec{x}(0)=\cv{1\\0.666}$ is close to $\cv{1\\ \frac{2}{3}}$.
   $\quad$Because $0.666$ not exactly equal to $\frac{2}{3}$ the solution curve deviates,
   $\quad$but stays in the beginning (and in the past) nearby.
   $\quad$The deviation from $\cv{1\\0.666}$ with $\cv{1\\ \frac{2}{3}}$ : $$\begin{aligned}\cv{1\\0.666}&=\cv{\frac{2}{3}\cdot \frac{1}{0.666}\\ \frac{2}{3}}= \cv{\frac{2}{1.998}\\ \frac{2}{3}}\\ &=\cv{\frac{2}{1.998}-1\\ 0}+\cv{1\\ \frac{2}{3}}\\ &=\cv{\frac{0.002}{1.998}\\ 0}+\cv{1\\ \frac{2}{3}}\\ &=\frac{2}{1998}\cdot\cv{1\\0}+ \cv{1\\ \frac{2}{3}}\\ &=\frac{1}{999}\cdot\cv{1\\0}+\frac{1}{3}\cdot\cv{3\\ 2}\end{aligned}$$ $\quad$ In the general solution (b) we have $\alpha=\frac{1}{999}$ and $\beta=\frac{1}{3}$ i.e. $$\vec{x}(t)= \frac{1}{999}e^{3t}\cdot\cv{1\\0}+\frac{1}{3}\beta e^{-t}\cdot\cv{3\\2}$$ $\quad$Then: $$\begin{aligned}\vec{x}(1) &= \frac{1}{999}e^{3}\cdot\cv{1\\0}+\frac{1}{3e}\cdot\cv{3\\2} \\ \\ \vec{x}(2) &= \frac{1}{999}e^{6}\cdot\cv{1\\0}+\frac{1}{3e^2}\cdot\cv{3\\2}\\ \\ \vec{x}(4) &= \frac{1}{999}e^{12}\cdot\cv{1\\0}+\frac{1}{3e^4}\cdot\cv{3\\2}\\ \\ \vec{x}(8) &= \frac{1}{999}e^{24}\cdot\cv{1\\0}+\frac{1}{3e^8}\cdot\cv{3\\2}\approx \cv{2.65\cdot 10^7\\ 0.006}\end{aligned}$$
6. $\quad\vec{x}(0)=\cv{1\\0.668}$ is close to $\cv{1\\ \frac{2}{3}}$.
   $\quad$Because $0.668$ the solution curve deviates,
   $\quad$but stays in the beginning (and in the past) nearby.
   $\quad$The deviation from $\cv{1\\0.668}$ with $\cv{1\\ \frac{2}{3}}$: $$\begin{aligned}\cv{1\\0.668}&=\cv{\frac{2}{3}\cdot \frac{1}{0.668}\\ \frac{2}{3}}= \cv{\frac{2}{2.004}\\ \frac{2}{3}}=\cv{\frac{1}{1.002}\\ \frac{2}{3}}\\ &=\cv{\frac{1}{1.002}-1\\ 0}+\cv{1\\ \frac{2}{3}}\\ &=\cv{\frac{-0.002}{1.002}\\ 0}+\cv{1\\ \frac{2}{3}}\\ &=-\frac{1}{501}\cdot\cv{1\\0}+\frac{1}{3}\cdot\cv{3\\ 2}\end{aligned}$$ $\quad$In the general solution (b) we have $\alpha=-\frac{1}{501}$ and $\beta=\frac{1}{3}$ i.e. $$\vec{x}(t)= -\frac{1}{501}e^{3t}\cdot\cv{1\\0}+\frac{1}{3}\beta e^{-t}\cdot\cv{3\\2}$$ $\quad$Then: $$\begin{aligned}\vec{x}(1) &= -\frac{1}{501}e^{3}\cdot\cv{1\\0}+\frac{1}{3e}\cdot\cv{3\\2} \\ \\ \vec{x}(2) &=- \frac{1}{501}e^{6}\cdot\cv{1\\0}+\frac{1}{3e^2}\cdot\cv{3\\2}\\ \\ \vec{x}(4) &= -\frac{1}{501}e^{12}\cdot\cv{1\\0}+\frac{1}{3e^4}\cdot\cv{3\\2}\\ \\ \vec{x}(8) &= -\frac{1}{501}e^{24}\cdot\cv{1\\0}+\frac{1}{3e^8}\cdot\cv{3\\2}\approx \cv{-5.29\cdot 10^7\\ 0.006}\end{aligned}$$
7. $\quad(0,0)$ is an equilibrium.
   $\quad$A solution to the line $x_2 = \frac{2}{3} x_1$ moves toward the equilibrium
   $\quad$A solution curve that starts close to the line $x_2=0$ moves away from the equilibrium.
   $\quad$A solution curve that starts close to the line $x_2=\frac{2}{3}x_1$ moves away from it and
   $\quad$goes to a horizontal axis.
   $\quad$The equilibrium $(0,0)$ is a saddle point.
   $\quad$This is also illustrated in the phase portrait below.
   $\quad$
   $\quad$Each solution satisfies the equation $(2x_1-3x_2)\cdot x_2^3=c$ for some constant $c$.