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

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The discussion focuses on finding equilibrium points and sketching the phase portrait for the given system of differential equations. The equilibrium points are identified at (5,5) and (-2,-2), which correspond to a spiral source and spiral sink, respectively. The Jacobian matrix is constructed, with its trace and determinant calculated to analyze stability. The user realizes that solving the system of equations directly simplifies the process. The conversation highlights the importance of understanding both analytical and graphical methods in studying dynamical systems.
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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