# 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

• See example 32.4; math.psu.edu/srikrish/math251/notes/realeigs.pdf Mar 21, 2015 at 8:59
• Oh Thanks the notes looks good. Btw how did u manage to find this set of notes? Is the notes for the entire course math251 available as well? Mar 21, 2015 at 13:27
• You can get the notes for this class at: math.psu.edu/srikrish/math251 Mar 21, 2015 at 13:44

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: