# Differential Equation stability at critical point

1. Mar 8, 2009

1. The problem statement, all variables and given/known data

This is the autonomous differential equation: x" - 2x' + 37x = 0

Solve the above DEQ and state whether the critical point (0,0) is stable, unstable, or semi-stable.

2. Relevant equations

Solution to the above DEQ is x = c1excos6x + c2exsin6x

3. The attempt at a solution

I worked out the solution using the quadratic formula and got roots 1$$\pm$$6i. This gives you an $$\alpha$$ of 1 and a $$\beta$$ of 6, which yields the equation I put in part 2 above.

From there, I read that when you get a general solution in the form x = e$$\alpha$$t(c1cos$$\beta$$t + c2sin$$\beta$$t) with $$\alpha$$ < 0 and $$\beta$$ $$\neq$$0, then you have a spiral point.

My problem is I'm not sure how to classify the stability of the critical point (0,0).

2. Mar 8, 2009

### HallsofIvy

It's easy. Your characteristic roots have positive real part so (0, 0) is an unstable critical point. You can see why from your general solution. (0,0) (in the phase plane) is a critical point because x(t)= 0 for all t (so that x'(t)= 0 also) is itself a solution. It is an unstable critical point because if x is just slightly different from 0, that exponential means it gets larger and larger, moving rapidly away from (0,0).