I am trying to solve for the nullclines of this system, but I know I am making a mistake because they aren't intersecting when they should. The system is: $x'=a-x-4xy/(1+x^2)$ $y'=bx(1-y/(1+x^2$))
When I tried to solve for them I got the x nullcline as $y=(x^2 +1)(a-x)/4x$ and the y nullcline as $y=1+x^2$. What mistake am I making? I think the x nullcline is fine, and the problem is with the y.
We are given:
$$x'=a-x-4xy/(1+x^2) \\ y'=bx(1-y/(1+x^2)$$
We want to find the nullclines (isoclines) for this system.
We can write:
$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{bx(1+x^2 - y)}{(a-x)(1+x^2) - 4xy}$$
Horizontal Slope
We want to find where $y' = 0$, so get:
$$y = (1+x^2), b = 0$$
Vertical Slope
We want to find where $x' = 0$, so get:
$$y = \dfrac{(a-x)(1+x^2)}{4x}, a= 0 ~~\mbox{and}~~ x = 0$$
Now, I am going to draw three different phase portraits and here is what you should do. Print them out, superimpose the isoclines onto them and see if you can figure out what happens on either side of the isocline (is the slope positive or negative). Superimpose both the $x$ and $y$ isoclines onto each phase portrait. Then, see if you can reason out the phase portrait curves based on all of the qualitative data.
Note, I chose what I think are three interesting cases based on the above, but more are possible, so you should explore.