Determining stability of the differential equation

Determine wether the differential equation is stable or not for the following conditions:

$x”'(0) = x”(0) = x'(0) = x(0) =0$

$\frac{d^3x(t)}{dt^3}+\frac{d^2x(t)}{dt^2}+4\frac{dx(t)}{dt}+4x(t) = e^t(H(t)-H(t-3))$

Now, I got that the solution to this differential equation reduces x(t) by use of Laplace Transforms, however, how do I determine whether or not it is stable?

$x(t) = 0.1e^{-t} + 0.1e^t – 0.1\sin(2t) + 0.1e^3 (\sin(2(t-3)) + e^{-t+3}-e^{t-3})*H(t-3)$

Now I first thought I would look at it graphically and I used Matlab software to get a plot as shown below:

And from the plot above it clearly appears to diverge (though this is not a proof).

Anyone have an idea as to an elegant proof to show the solution is unstable?

Anyone help would be greatly appreciated!

Solutions Collecting From Web of "Determining stability of the differential equation"

You have eigenvalues $-1,2i,-2i$ and no forcing after $t=3$. This is stable, but not asymptotically stable, i.e., solutions do not shrink to zero, but they also do not grow.

As you found, the homogeneous solution, which is valid for $t>3$, is
$$x(t)=c_1\cos(2t)+c_2\sin(2t)+c_3e^{-t}$$
where the first two components stay bounded while the third converges towards zero.

On the graph

That a formula represents a stable solution does not automatically mean that the evaluation of that formula is numerically stable. Indeed, for large $t$ the floating point evaluation of that formula reduces to the evaluation of $0.1·(e^t-e^3·e^{t-3})$ which can incur cancellation error of the size $0.1·e^t·10^{-16}$ which for $t=100$ gives the magnitude $10^{-1+43-16}=10^{26}$ which is indeed also the scale factor in the graph.

One needs to explicitly switch off this exponential factor via
$$0.1·e^t·(H(t)-H(t-3))$$
or a similar form of conditional evaluation.