We have the following system of ordinary differential equation. To solve this using diagonlsation we do the following. My concern is that my general solution is a subtly different result, I've checked with a matrix calculator and mine seems to be correct? But the subtle difference really goes on to affect the analysis I think.
- $\frac{dX}{dt} = -cY$
- $\frac{dY}{dt} = -cX$
$\begin{bmatrix} \frac{dX}{dt} \\ \frac{dY}{dt} \end{bmatrix} = \begin{bmatrix} 0 & -c \\ -c & 0 \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}$ Let's denote the matrix $A =\begin{bmatrix} 0 & -c \\ -c & 0 \end{bmatrix}$
$\det \begin{bmatrix} -\lambda & -c \\ -c & -\lambda \end{bmatrix} = \lambda^2 - c^2 = 0$
The eigenvalues are: $\lambda_1 = c, \quad \lambda_2 = -c$
- $\lambda_1 = c$: $(A - cI)v_1 = 0$
$\begin{bmatrix} -c & -c \\ -c & -c \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0$
$v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$
- $\lambda_2 = -c$: $(A + cI)v_2 = 0$
$\begin{bmatrix} c & -c \\ -c & c \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0$
$v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$
This gives me the general solution:
$X(t) = -\alpha e^{ct} + \beta e^{-ct}$
$Y(t) = \alpha e^{ct} + \beta e^{-ct}$
But what they get is
$X(t) = \alpha e^{ct} + \beta e^{-ct}$
$Y(t) = -\alpha e^{ct} + \beta e^{-ct}$
This then goes on to affect my analysis, for instance it turns out that $\alpha < 0$ therefore $-\alpha = |\alpha|$ which matters which equation it's in.
Note: They used substitution, and i wonder if that's how they got the different answer? But this would highlight exactly the concerns i had in a previous post, regarding the different general results for the two methods, particularly in say an exam setting.
Note: Technically these were "Lanchester Equations" - but this seems irrelevant to the analysis.