Use two solutions to a high order linear homogeneous differential equation with constant coefficients to say something about the order of the DE OK, this one utterly baffles me.
I am given two solutions to an nth-order homogeneous differential equation with constant coefficients. Using the solutions, I am supposed to put a restriction on n (such as n>=5)
I have no idea what method, theorem, or definition is useful to do this. 
My current "theory" is that I must find all the different derivatives of the solutions and tally up how many unique derivatives they have. This is wrong, but am I going in the right direction?
The specific solutions for the example are t^3 and (t)(e^t)(sint)
These solutions are to an nth-order homogeneous differential equation with constant coefficients, which means that n >= ?
Thanks in advance.
 A: The roots of your ODE can be written as $t^3e^{0\times t},te^t\sin{t}$. So, the characteristic equation has a root at $0$ of multiplicity at least $4$ and a root at $1+i$ of multiplicity at least $2$. If your coefficients are allowed to be complex. the minimum degree is $6$. If they are real, it will also have conjugate roots at $1-i$ of multiplicity equal to that of root at  $1+i$. In this case, minimum degree  $=4+2+2 =8$.
UPDATE: For complex coefficients, the characteristic function in the minimal case (degree 6) is $x^4(x-1-i)^2$. For real coefficients, it is $x^4(x-1-i)^2(x-1+i)^2 = (x^4(x^2-2x+2)^2$ (degree 8)
A: A related problem. We will use the annihilator method. Note that, since you are given two solutions of the ode with constant coefficients, then their linear combination is a solution to the ode too. This means the function
$$ y(x) = c_1 t^3 + c_2 te^{t}\sin(t) $$
satisfies the ode. Applying the operator $D^4((D-1)^2+1)^2,$ where $D=\frac{d}{dx},$ to the above equation gives
$$D^4((D-1)^2+1)^2 y(x) = 0.$$
From the left hand side of the above equation, one can see that the differential equation is at least of degree $8$ or $n\geq 8.$
