What is the minimal order of a homogeneous linear ODE with constant coefficients that admits the solution $$y(t)=t^2e^{-t/3}cos(2t) ?$$

What I tried:

My guess is that it is a third order. Since $t^2e^{-t/3}cos(2t)$ is a solution, then $te^{-t/3}cos(2t)$ and $e^{-t/3}cos(2t)$ also have to be the other solutions with $t^2e^{-t/3}sin(2t)$, $te^{-t/3}sin(2t)$, and $e^{-t/3}sin(2t)$ being the conjugate pairs of the solutions respectively. However what I'm confused about is whether the minimal is third or sixth order. Do I have to take into consideration the conjugate pair when I'm evaluating the minimal order of ODE? Could anyone please explain?


Hint Since the o.d.e. is homogeneous and $u(t) := t^2 e^{-t / 3} \cos 2 t$ is a solution, so are the functions $u_k := u(t - \frac{k \pi}{4})$, $k = 0, 1, 2, \ldots$.

But checking (say, by computing the Wronskian) shows that the solutions $u_k, k=0, \ldots, 5$ are linearly independent, so the o.d.e. has order at least $6$.

On the other hand, one can readily construct a homogeneous, order $6$ o.d.e. with constant coefficients and with solution $u$: The form of the given solution $u$ suggests the characteristic polynomial of any such o.d.e. must have triple roots at each of $-\frac{1}{3} \pm 2i$, which gives the order $6$ polynomial $$p(r) = \left[r - \left(-\frac{1}{3} + 2i\right)\right]^3 \left[r - \left(-\frac{1}{3} - 2i\right)\right]^3 = \left(r^2 - \frac{2}{3} r + \frac{37}{9}\right)^3,$$ and by construction the corresponding (somewhat messy) o.d.e. has $u(t)$ as a solution.


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