Plotting the graph of systems of ODE The eigenvalues and eigenvectors of a matrix A are given.
Consider the corresponding system $x' = Ax$.
(a) Sketch a phase portrait of the system.
(b) Sketch the trajectory passing through the initial point $(2, 3)$.
(c) For the trajectory in part (b) sketch the graphs of $x_1$ versus t and of $x_2$ versus t on the same set of axes.
$$r_1=1,n_1 = \left(  
\begin{array}{c}
     1\\
    2\\
  \end{array}
\right)$$
$$r_2=2,n_2 = \left(  
\begin{array}{c}
     1\\
    -2\\
  \end{array}
\right)$$
Where $r$ represents the eigenvalues and $n$ represents the eigenvectors.
What i tried
Since both eigenvalues are positive, the plot must be a nodal source with unstable equilibrium. Hence the graph points away from the origin. From what i know,  Eigenvalues that are negative will correspond to solutions that will move towards the origin as t increases in a direction that is parallel to its eigenvector.  Likewise, eigenvalues that are positive move away from the origin as t increases in a direction that will be parallel to its eigenvector.However im unsure of how to intepret the eigenvectors to plot the graph. Could anyone please explain. Thanks
 A: To visualize the solution, it helps to graph it in the $x_1 x_2$-plane for various $c_1, c_2$.
I will map out most of the details and you can add the rest. Start with the first eigenvalue/vector solution and we have in scalar form:
$$x_1 = c_1 e^t, x_2 =2 c_1 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = 2 x_1$. If $c_1 \gt 0$, we are in quadrant $I$ and if $c_1 \lt 0$, we are in quadrant $III$.
Repeating this for the second eigenvalue/vector, we have:
$$x_1 = c_2 e^t, x_2 =-2 c_2 e^{2t}$$
By eliminating $t$ between these two equations, we get $x_2 = -2 x_1$. If $c_2 \gt 0$, we are in quadrant $IV$ and if $c_2 \lt 0$, we are in quadrant $II$.
The solution is a linear combination of these two as:
$$x(t) = c_1 e^{t} \left( \begin{array}{c}   1\\  2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c}  1\\ -2\\  \end{array}\right)$$
So, we can plot the eigenvectors as (blue is $n_1$ and yellow is $n_2$):

Now, what happens to the solutions as $t \rightarrow + \infty$? Just draw handfuls of those around the lines in each quadrant (hint - they are asymptotic to the line and move away from the origin as this is an unstable node). This will give you a phase portrait that is an unstable node.
For part $b.$, we have the initial point $(2, 3)$. Since we know we are going to be asymptotic to the eigenvector in quadrant $I$, it starts at ($2, 3)$ and goes to this line.
For part $c.$, we have $x(0) = (2, 3)$, with:
$$x(t) = c_1 e^{t} \left( \begin{array}{c}   1\\  2\\ \end{array}\right) + c_2 e^{2 t} \left(\begin{array}{c}  1\\ -2\\  \end{array}\right)$$
We end up with $c_1 = \dfrac 74, c_2 = \dfrac 14$, so have:
$$x_1(t) = \frac{7}{4} e^t+\frac{1}{4} e^{2t}, x_2(t) = \frac{7}{2}e^{t}-\frac{1}{2} e^{2t}$$
A plot of this shows:

One last note, it helps to plot this parametrically and we get:

