I am trying to get a linearised matrix system of the following non-linearised ODE system, with two degree of freedom. My approach has been using Taylor expansion for the $B + x_2^C$ denominator but I get an even messier expression instead...
Any guidance on how I should re-tackle this?
$$ \dfrac{dx_1}{dt} = \dfrac{A}{B + x_2^C} - Dx_1 $$
$$ \dfrac{dx_2}{dt} = E x_1 - Fx_2 $$
A->F are just constants. $x_2^C$ is just $x_2$ to the power of C.
The linearisation of a system at a certain point is given by the Jacobian matrix of (the right hand side of) that system, evaluated at that certain point. In your case, you have a system of the form \begin{align} \frac{\text{d} x_1}{\text{d} t} &= f(x_1,x_2),\\ \frac{\text{d} x_2}{\text{d} t} &= g(x_1,x_2). \end{align} The Jacobian matrix at the point $(a,b)$ is therefore given by \begin{equation} J = \begin{pmatrix} \frac{\partial f}{\partial x_1}(a,b) & \frac{\partial f}{\partial x_1}(a,b) \\ \frac{\partial g}{\partial x_1}(a,b) & \frac{\partial g}{\partial x_2}(a,b) \end{pmatrix}. \end{equation} In your case, the only difficult partial derivative is $\frac{\partial f}{\partial x_2}$, but using the chain rule, you should be able to calculate that.