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).
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.