- #1
FeDeX_LaTeX
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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?
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?