# Equilibrium Points of DE

1. Oct 27, 2013

### FeDeX_LaTeX

The problem statement, all variables and given/known data
Find the equilibrium points of the system, determine their type and sketch the phase portrait.

$\frac{dx}{dt} = -3y + xy - 10, \frac{dy}{dt} = y^2 - x^2$

The attempt at a solution

Putting it together:

$\frac{dy}{dx} = \frac{y^2 - x^2}{-3y + xy - 10} \equiv \frac{Q(x,y)}{P(x,y)}$

Here, we see that the horizontal nullclines are plotted along the line $y = \pm x$ and the vertical nullclines along the curve $y = \frac{10}{x - 3}$.

We form the Jacobian, i.e.

J = $\left( \begin{array}{cc} P_x & P_y \\ Q_x & Q_y \end{array} \right)$ = $\left( \begin{array}{cc} y & x - 3 \\ -2x & -2y \end{array} \right)$

So $-tr(J) = y$ and $det(J) = 2x^2 - 2y^2 - 3$.

My question is, where do I go from here? Through using a differential equation plotter, I can see that the equilibrium points are a spiral source and spiral sink at (5,5) and (-2,-2) respectively. How does one deduce this from the Jacobian?

2. Oct 27, 2013

### FeDeX_LaTeX

Never mind, I've overcomplicated it -- all I needed to do was solve that system of DEs for x and y (substituting x = y).

The magic of the Homework board strikes again!