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Given a differential equation, for example, a logistic curve, how do I determine the equilibrium points, graphically?

Consider $$x'=ax\left(1-\frac{x}{b}\right)-\frac{x^2}{1+x^2}$$

It is clear that $x=0$, is one, how do I determine/demonstrate other equilibrium given that I do not have values of the $a$ and $b$ as well. Is it more possible if I have a value of one of the parameters? I also would like to classify equilibrium points as stable and unstable.

Are there special kinds of plots for these?

I am looking for, preferable, graphical solution.


So, if $x\neq 0$, then $$a\left(1-\frac{x}{b}\right)=\frac{x}{1+x^2}$$

How do I solve this graphically?

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The equilibrium point is defined as $\dot x=0$. What do you mean by other equilibrium? – Occupy Gezi May 3 '13 at 22:31

Equilibrium points are where the rate of change is zero. You have an equation for the first derivative which represents the rate of change. Plot it, where it crosses the axis are your equilibrium points.

Further, if you can infer info about the stability of the equilibrium. For it to be stable, it must be >0 to the left and <0 to the right. Do you see why?

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You need to solve $$a\left(1-\frac{x}{b}\right)=\frac{x}{1+x^2}$$ for non-zero equilibrium solutions. Since $x\neq 0$ and $1+x^2>0$, this can be rearranged to $$\frac{a}{b}x^3-ax^2+\left(\frac{a}{b}+1\right)x+a=0.$$

Given that the question is tagged as matlab, I assume that you need to find the equilibrium points in MATLAB. This can be done as follows:

% given a and b

coeff = [a/b,-a,a/b+1,a];
equib_values = roots(coeff);

The equilibrium values are then given in equib_values.

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