# 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 >= ?

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Are you sure you got the roots correct? I think they should be $e^{3t}$ adn $te^t\sin{t}$. Homogeneous ODE have always exponents as roots –  dexter04 Dec 11 '12 at 18:08
@dexter04 $t^3 = t^3\exp(0\cdot t)$ is ok. –  martini Dec 11 '12 at 18:10
Here's a copy-paste of the (true or false) question: If an nth order linear homogeneous differential equation with constant coefficients has (t^3) and (t)(e^t)(sint) as solutions, then n >= 8 –  Marlin Dec 11 '12 at 18:13

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 $n\leq 8.$

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Unless I am mistaken, you are saying that the minimum order of the ODE must be equal to the highest degree of D in the expanded operator. I must learn more about this method. Thank you very much Mhenni, you are a savior. –  Marlin Dec 11 '12 at 18:47
@Marlin: Yes, because maybe the ode has other solutions. We have been given only two. –  Mhenni Benghorbal Dec 11 '12 at 18:53
@Marlin: You are welcome. –  Mhenni Benghorbal Dec 11 '12 at 18:53
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)
Exactly!! For a solution of the form $t^ke^{\alpha t}$, the characteristic equation has a $k+1$ roots at $\alpha$. –  dexter04 Dec 11 '12 at 19:25