# Thread: Non-equilibrium solutions of an ODE

1. ## Non-equilibrium solutions of an ODE

So the equilibrium solutions are -1 and 1, but what are the non-equilibrium solutions?

I started with:

$dy/(1-y^2) = dt$

$-ln(1-y^2) = t + c$

Am I going in the right direction?

2. Your integration of the DE seems a bit off, to me. Try the $y$ integration again.

3. Unfortunately not. Remember, the integral of $dU/U$ is the natural log of the absolute value of U, but you don't have the differential of your denominator, $-(y^2-1)$ in the numerator. You'll have to use partial fraction decomposition.

4. A simple preliminary analysis of the DE...

$y^{'} = 1 - y^{2}$ (1)

... permits You to conclude that is...

a) $y^{'} >0$ for $-1< y < 1$

b) $y^{'} = 0$ for $y=1$ or $y=-1$

c) $y^{'} < 0$ for $y=1$ or $y=-1$

... so that is...

a) $\lim_{t \rightarrow \infty} y(t) = 1$ for $y(0) >-1$

b) $\lim_{t \rightarrow \infty} y(t) = - 1$ for $y(0) = -1$

c) $\lim_{t \rightarrow \infty} y(t) = -\infty$ for $y(0) <-1$

Kind regards

$\chi$ $\sigma$