Problem Statement
A closed chemical process involves two reactants with concentrations $$x(t)$$ and $$y(t)$$ governed by:\n\n$$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 1 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}, \quad x(0) = 4, \, y(0) = 1$$\n\n(a) Find the eigenvalues and eigenvectors of the coefficient matrix. [4 marks]\n\n(b) Write down the particular solution for $$x(t)$$ and $$y(t)$$. [3 marks]\n\n(c) Classify the nature of the equilibrium at $$(0, 0)$$. [1 mark]
Verified Solution & Marking Scheme
Part (a): Characteristic Equation
$$\det(A - \lambda I) = (-4-\lambda)(-3-\lambda) - 2 = \lambda^2 + 7\lambda + 12 - 2 = \lambda^2 + 7\lambda + 10 = 0 \implies (\lambda + 2)(\lambda + 5) = 0$$.\nEigenvalues: $$\lambda_1 = -2, \quad \lambda_2 = -5$$.
Find Eigenvectors
For $$\lambda_1 = -2$$: $$\begin{pmatrix} -2 & 2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = v_2 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$.\nFor $$\lambda_2 = -5$$: $$\begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0 \implies v_1 = -2v_2 \implies \mathbf{v}_2 = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$.
Part (b): Particular Solution
General: $$\begin{pmatrix} x \\ y \end{pmatrix} = c_1 e^{-2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 e^{-5t} \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$.\nAt $$t = 0$$: $$c_1 - 2c_2 = 4$$ and $$c_1 + c_2 = 1$$. Subtracting gives $$-3c_2 = 3 \implies c_2 = -1, \, c_1 = 2$$.\n$$x(t) = 2e^{-2t} + 2e^{-5t}, \quad y(t) = 2e^{-2t} - e^{-5t}$$.
Part (c): Classification
Both eigenvalues are strictly negative and real ($$\lambda_1 = -2, \, \lambda_2 = -5$$). The equilibrium is a stable node (sink).