Finding Equilibrium Points and Sketching Phase Portrait for a System of DEs

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SUMMARY

The discussion focuses on finding equilibrium points and sketching the phase portrait for the system of differential equations defined by the equations dx/dt = -3y + xy - 10 and dy/dt = y^2 - x^2. The equilibrium points identified are (5,5) as a spiral source and (-2,-2) as a spiral sink. The Jacobian matrix is constructed to analyze the stability of these points, with the trace and determinant calculated as -tr(J) = y and det(J) = 2x^2 - 2y^2 - 3. The solution emphasizes the importance of solving the system of DEs directly to simplify the analysis.

PREREQUISITES
  • Understanding of differential equations and their stability analysis
  • Familiarity with Jacobian matrices and their role in determining equilibrium point types
  • Knowledge of phase portraits and nullclines in dynamical systems
  • Experience with differential equation plotters for visual analysis
NEXT STEPS
  • Study the construction and interpretation of Jacobian matrices in dynamical systems
  • Learn how to analyze stability using eigenvalues of the Jacobian
  • Explore phase portrait sketching techniques for nonlinear systems
  • Investigate the use of software tools like MATLAB or Python's Matplotlib for plotting differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, dynamical systems, and stability analysis. This discussion is also beneficial for anyone looking to enhance their skills in visualizing and interpreting phase portraits.

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Homework Statement
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?
 
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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!
 

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