Problem Statement
A coupled dynamical system is described by the linear system of differential equations:
$$\frac{dx}{dt} = 3x - 2y$$
$$\frac{dy}{dt} = 2x - 2y$$
with initial conditions $x(0) = 3$ and $y(0) = 2$.
(a) Show that the eigenvalues of the system matrix $A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}$ are $\lambda_1 = 2$ and $\lambda_2 = -1$. [3 marks]
(b) Find corresponding eigenvectors for $\lambda_1$ and $\lambda_2$. [3 marks]
(c) Hence, determine the unique exact solution for $x(t)$ and $y(t)$. [4 marks]
Verified Solution & Marking Scheme
Part (a): Characteristic equation of A
$$\det(A - \lambda I) = \begin{vmatrix} 3 - \lambda & -2 \\ 2 & -2 - \lambda \end{vmatrix} = (3 - \lambda)(-2 - \lambda) - (-4)$$
$$= -6 - 3\lambda + 2\lambda + \lambda^2 + 4 = \lambda^2 - \lambda - 2 = 0$$
Factoring: $$(\lambda - 2)(\lambda + 1) = 0 \implies \lambda_1 = 2, \, \lambda_2 = -1 \quad \text{(AG)}$$
Part (b): Determine eigenvectors
For $\lambda_1 = 2$: $(A - 2I)\vec{v}_1 = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix}\begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Equation: $v_{11} - 2v_{12} = 0 \implies v_{11} = 2v_{12}$. Choosing $v_{12} = 1$, eigenvector $\vec{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.
For $\lambda_2 = -1$: $(A + I)\vec{v}_2 = \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Equation: $2v_{21} - v_{22} = 0 \implies v_{22} = 2v_{21}$. Choosing $v_{21} = 1$, eigenvector $\vec{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
Part (c): Form general solution and apply initial conditions
General solution:
$$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix} = c_1 e^{2t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
At $t = 0$:
$$2c_1 + c_2 = 3$$
$$c_1 + 2c_2 = 2$$
Multiply second equation by 2: $2c_1 + 4c_2 = 4$.
Subtracting: $3c_2 = 1 \implies c_2 = \frac{1}{3}$.
Then $c_1 = 2 - 2(1/3) = \frac{4}{3}$.
Write individual explicit functions
$$x(t) = 2c_1 e^{2t} + c_2 e^{-t} = \frac{8}{3}e^{2t} + \frac{1}{3}e^{-t}$$
$$y(t) = c_1 e^{2t} + 2c_2 e^{-t} = \frac{4}{3}e^{2t} + \frac{2}{3}e^{-t}$$