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.

  • $\begingroup$ 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 $\endgroup$ – dexter04 Dec 11 '12 at 18:08
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    $\begingroup$ @dexter04 $t^3 = t^3\exp(0\cdot t)$ is ok. $\endgroup$ – martini Dec 11 '12 at 18:10
  • $\begingroup$ 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 $\endgroup$ – 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 $n\geq 8.$

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    $\begingroup$ 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. $\endgroup$ – Marlin Dec 11 '12 at 18:47
  • $\begingroup$ @Marlin: Yes, because maybe the ode has other solutions. We have been given only two. $\endgroup$ – Mhenni Benghorbal Dec 11 '12 at 18:53
  • $\begingroup$ @Marlin: You are welcome. $\endgroup$ – 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)

  • $\begingroup$ Please tell me if I have this right: You get multiplicity = 4 from all the (t) terms, which are t^3 and t. If complex values are allowed, I can make the characteristic equation equal 1+i as well as 0 which adds 2 to the multiplicity. I'm losing my grip on what the characteristic equation would look like. However in this case, complex values are not allowed so I must cancel the 1+i by making the characteristic equation equal 1-i, giving the multiplicity a final value of 8. Any chance you could write the characteristic equations? Thanks for the reply btw $\endgroup$ – Marlin Dec 11 '12 at 18:57
  • $\begingroup$ Sorry to bother you again, but I am having difficulty making initial characteristic equations using the given solutions. EDIT: Is it (x-a)(x-b) with a and b coming from the exponential functions in the solutions? $\endgroup$ – Marlin Dec 11 '12 at 19:19
  • $\begingroup$ Exactly!! For a solution of the form $t^ke^{\alpha t}$, the characteristic equation has a $k+1$ roots at $\alpha$. $\endgroup$ – dexter04 Dec 11 '12 at 19:25
  • $\begingroup$ Ok, so all the x^4 comes from the left solution (t^3). The right solution has k=1 and a=1, meaning k+1 = 2 roots that make it equal a (or 1). The complex roots come from the sin(t) working backwards from Euler's formula. Does it matter if I put (1-i) into the characteristic equation before (1+i)? $\endgroup$ – Marlin Dec 11 '12 at 19:45
  • $\begingroup$ No, ordering of roots is unimportant. $\endgroup$ – dexter04 Dec 11 '12 at 20:11

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