Differential Equation stability at critical point

[JJBladester]

Homework Statement

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.

Homework Equations

Solution to the above DEQ is x = c₁e^xcos6x + c₂e^xsin6x

The Attempt at a Solution

I worked out the solution using the quadratic formula and got roots 1\pm6i. 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^\alphat(c₁cos\betat + c₂sin\betat) with \alpha < 0 and \beta \neq0, then you have a spiral point.

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

 

[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).