# 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.