# Phase Plane Analysis

Classify the fixed point at the origin and sketch an accurate phase portrait for the following system: $$\left\{\begin{matrix} \dfrac{dx}{dt}=36x-16y\\ \dfrac{dy}{dx}=-3x+28y \end{matrix}\right.$$

Am I correct in thinking that I need to write these two equations in matrix form, find the eigenvalues and depending on what they are will determine what fixed point I have? To sketch the phase portrait do I then need to know the eigenvectors and directions?

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Welcome to MSE. Are you sure that the second equation is $dy/dx$ and not $dy/dt$? –  Romeo Aug 12 '12 at 12:28
That's what I'm confused about! The question is definitely dy/dx. If I had dy/dt instead I would have no problem doing this question. –  Hobgoblin Aug 12 '12 at 12:36
It is probably a misprint. Otherwise, you'd end up with a nonlinear system. –  Siminore Aug 12 '12 at 12:41
If I take it as dy/dt instead of dy/dx and find the eigenvalues and eigenvectors do I need to know the eigendirections to draw the phase portrait? How would I go about finding the eigendirection? –  Hobgoblin Aug 12 '12 at 12:46
The eigendirections are given by the eigenvectors. Put differently, all the eigenvectors for a given eigenvalue form a linear space (at least if you add the zero vector, which is technically not an eigenvector, but still satisfies the defining equation of eigenvectors), and you can think of this space, which is typically one-dimensional, as specifying a direction. And that direction is the eigendirection. –  Harald Hanche-Olsen Aug 12 '12 at 13:48

Here we go:

y1' = dx/dt; y2' = dy/dt : Doesn't matter what;
y_vec' = [y1';y2']; y_vec = [y1;y2];
A = [36 -16;-3 28];
y_vec' = A*Y_vec;
p = a11+a22 = 36+28 = 64;
q = a11*a22 - a12*a21 = 36*28 - -16*-3 = 960;
delta = p^2 - 4q = 256;
therefore : improper node;

e_vec =
e_vec1    e_vec2
0.9701    0.8000
-0.2425    0.6000

eigenvalues =

40
24

therefore 2 straightlines are y2 = (0.9701/-0.2426)*y1; y2 = (0.8/-0.6)*y1;

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