# On constant-coefficient linear homogenous ODE [duplicate]

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).

• Mentionned it and explained the difference in my opinion Commented Jul 19, 2018 at 21:50
• I see, I agree it is not a proof so I'll delete that msg but keep in mind that understanding this can help you solve the problem, after proving uniqueness I am sure it can help.
– ℋolo
Commented Jul 19, 2018 at 21:54
• Already brought up, already explained why it isn't. Commented Jul 20, 2018 at 8:12

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.

• 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. Commented Jul 19, 2018 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. Commented Jul 19, 2018 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. Commented Jul 19, 2018 at 22:05
• @Matrefeytontias Uniqeness??? Commented Jul 19, 2018 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). Commented Jul 19, 2018 at 22:12