# 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:

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

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

Am I going in the right direction?

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

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

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

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

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

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

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

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

... so that is...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$