0

I haven't done any "serious" math in a while and I wanted to get back to it via differential equations. I remember very well the different methods to solve various sorts of linear ODEs, but I can't remember or even find online any information about the proofs of these methods. I'm starting with constant-coefficient linear homogenous ODEs with no Cauchy or boundary condition, and I remember that the solution space of these equations has the same dimensionality as the degree of the ODE. I remember that linear algebra is involved, and I have looked around on the internet for things about differential operators, to no avail (paper didn't get me very far either).

tl;dr : any hints for proving that the solution space of an nth degree constant-coefficient linear homogenous ODE has dimensionality n ? I know it's at least n because I can show n linearly independent solutions (the well-known exponentials).

EDIT : it has been mentionned that this question is similar to this one. Where I think they differ, in addition to pertaining to different specialisations of ODEs, is that this question asks for an explanation, while I'm looking for a proof. I know the result is true, I just don't remember how to prove it (as a side note, the accepted answer for the aforementionned question does not answer mine).

Matrefeytontias
  • 792

1 Answers1

0

Let$$ y^{(n)} = a_ny^{(n-1)}+...+a_3y''+a_2y'+a_1y+a_0$$

The characteristic equation is a polynomial $$ P(\lambda ) = \lambda^{(n)}- a_n\lambda^{(n-1)} -....-a_1$$

You will find $n$ linearly independent solutions for your differential equations because you have $n$ roots counting multiplicities for P(n) called eigenvalues.

For each simple eigenvalue you find a solution $e^{\lambda t}.$ For eigenvalues with multiplicities you find solutions $$ e^{\lambda t}, te^{\lambda t},...,t^ke^{\lambda t}.$$

The general solution is then found by a linear combinations of these solutions plus a particular solution.

Thus the solution set is an n-dimensional vector space.

Mohammad Riazi-Kermani
  • 69,891
  • As I said, I know that the linear combination of exponentials are solutions, hence why I said that I know that the solution space is at least n-dimensional. What I don't know, and which is missing from your answer as well (unless I missed it), is that since the characteristic equation itself is given by assuming that the solutions have the form $C t^k e^{\lambda t}$, how do we know that this assumption doesn't restrict the space of solutions found ? My idea was initially to show that $dim(ker(L)) = n$, where $L$ is the differential operator pertaining to the equation. – Matrefeytontias Jul 19 '18 at 21:41
  • @Matrefeytontias The general solution covers every solution. So you have all your solutions covered by the linear independent of these n solutions. You proved this by solving a system whose determinant is the Wronskian of these linearly independent exponential solutions. – Mohammad Riazi-Kermani Jul 19 '18 at 21:52
  • I think there is a misunderstanding. I do know that all the n solutions are linearly independent (as you said, a non-zero wronskian shows it) ; what I'm asking is, how do we know that there are not more solutions ? You do say that, and I quote, "the general solution covers every solution", but the very question I'm asking to begin with is "how do I prove that". I want to prove the uniqueness of these solutions, ie the converse of what you said ("these are solutions" vs "every solution is this"). Sorry if I'm explaining poorly. – Matrefeytontias Jul 19 '18 at 22:05
  • @Matrefeytontias Uniqeness??? – Mohammad Riazi-Kermani Jul 19 '18 at 22:09
  • Yes. You showed me $n$ linearly independent solutions, stating that the solution space is then $n$-dimensional, while as far as I know it only shows that it is at least $n$-dimensional. What I'm trying to prove is that there is no other solution outside of this $n$-dimensional space that you showed me (which, again, I already knew about). – Matrefeytontias Jul 19 '18 at 22:12