I am trying to linearize the following nonlinear system to determine the type of equilibrium point present. The system is
$x'=a-x-(4xy/(1+x^2))$
$y'=bx(1-(y/(1+x^2)))$
Do I do this by creating the jacobian, evaluating at the c.p., and solving for the eigenvalues? With the two parameters a,b things get really messy when I try to solve for the eigenvalues. I thought there was a simpler way by simply ignoring the nonlinear terms...
You have a differential equation of the form $$q' = f(q)$$ where $f$ is some nonlinear function. I assume you are starting from some known equilibrium point $q_0$ where $f(q_0) = 0$. Now suppose you are a small displacement $q(t) = q_0 + \delta q(t)$ away from $q_0$. Plugging in you get $$\delta q' = f(q_0 + \delta q).$$
Apply Taylor's theorem on the right-hand side to get $$\delta q' = f(q_0) + \nabla f \cdot \delta q + O(\delta q^2).$$ Assuming small displacements, and that the Taylor coefficients of $f$ decay sufficiently quickly, etc. we can ignore the higher-order terms, to get the approximate, linearized ODE $$\delta q' = 0 + \nabla f \cdot \delta q.$$
Solve this in the usual way, i.e. find the eigenvectors $v_i$ of $\nabla f$ with eigenvalues $\lambda_i$, and notice these give you solutions of the form $$\delta q = Ae^{\lambda_i t}v_i.$$