Originally Posted by

**jschlarb** Okay, I have the following autonomous differential equation:

$\displaystyle dx/dt = f(x) = x^2-5x+6$

I found the steady states of the equation of be x*1=2 and x*2=3

When I sub them into $\displaystyle f'(x^*) = 2x-5$, I get $\displaystyle f'(x^*1) = -1 $ and $\displaystyle f'(x^*2)= 1$.

Now I was told that the stability theorem states that if $\displaystyle f'(x^*) < 0$, then x* is stable and when $\displaystyle f'(x^*) > 0$, x* is unstable, however, when I create my phase line diagram is shows that x*2 is stable.

Could someone shed some light on this?

Thanks,

JS