91Ó°ÊÓ

A critical point of a system of differential equations \( \frac{dx}{dt}, \frac{dy}{dt} \) is found by setting the equations equal to zero: \( x - 5y - 5 = 0 \) and \( x - y - 3 = 0 \). Solve these equations simultaneously to find values of \( x \) and \( y \) that satisfy both equations. Start with the second equation: \( x = y + 3 \). Substitute into the first equation: \( (y + 3) - 5y - 5 = 0 \). Simplifying: \( -4y - 2 = 0 \) gives \( y = -0.5 \). Using \( x = y + 3 \), we find \( x = 2.5 \). The critical point is therefore \( (2.5, -0.5) \).

To classify the critical point, we need to linearize the system around \( (2.5, -0.5) \). The linearization involves finding the Jacobian matrix of the system's vector field. The original system is \[ \begin{pmatrix} \dot{x} \ \dot{y} \end{pmatrix} = \begin{pmatrix} x - 5y - 5 \ x - y - 3 \end{pmatrix}. \] The Jacobian matrix \( J \) is formed by taking partial derivatives with respect to \( x \) and \( y \), resulting in: \[ J = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & -5 \ 1 & -1 \end{pmatrix}. \]

The eigenvalues of the Jacobian matrix \( J \) determine the type and stability of the critical point. Start by finding these eigenvalues by solving the characteristic equation \( \det(J - \lambda I) = 0 \). Substituting the values, we obtain: \[ \det \begin{pmatrix} 1 - \lambda & -5 \ 1 & -1 - \lambda \end{pmatrix} = 0. \] This simplifies to \( (1 - \lambda)(-1 - \lambda) + 5 = \lambda^2 - 1\lambda + \lambda + 5 = \lambda^2 + \lambda + 4 = 0 \). Solve for \( \lambda \) using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = 1, c = 4 \), giving eigenvalues \( \lambda = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 4}}{2 \times 1} = \frac{-1 \pm \sqrt{-15}}{2} \). The eigenvalues are complex \( \lambda = -0.5 \pm i \sqrt{3.75} \).

The critical point is classified based on the nature of its eigenvalues. Since the eigenvalues have a negative real part and are complex, \( \lambda = -0.5 \pm i \sqrt{3.75} \), the critical point \((2.5, -0.5)\) is a spiral sink. This indicates that nearby trajectories will spiral into the critical point, indicating stability.

Use a computer algebra system or graphing calculator to plot the phase portrait. Input the differential equations \( \dot{x} = x - 5y - 5 \) and \( \dot{y} = x - y - 3 \). Observe the behavior of trajectories in the phase space. The trajectories should converge to the point \((2.5, -0.5)\), forming a spiral pattern around it, confirming the presence of a stable spiral sink.