Problem Statement
Consider the linear coupled differential system:
$$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
(a) Show that the eigenvalues of the coefficient matrix are both zero. [3 marks]
(b) Find the line of equilibrium points for this system. [3 marks]
(c) Describe the nature of the phase portrait trajectories. [3 marks]
Verified Solution & Marking Scheme
Part (a): Characteristic equation
$$\det(\mathbf{A} - \lambda\mathbf{I}) = (2 - \lambda)(-2 - \lambda) - (1)(-4) = -4 + \lambda^2 + 4 = \lambda^2 = 0 \implies \lambda_1 = \lambda_2 = 0$$
Part (b): Equilibrium line
Setting $$\mathbf{A}\mathbf{x} = \mathbf{0} \implies 2x + y = 0 \implies y = -2x$$. Every point on the line $$y = -2x$$ is an equilibrium point.
Part (c): Trajectory directions
Since $$\dot{x} = 2x + y$$ and $$\dot{y} = -2(2x + y) = -2\dot{x}$$, we have $$\frac{dy}{dx} = -2$$. All trajectories are parallel straight lines with slope $$-2$$ moving toward or away from the equilibrium line $$y = -2x$$.