# Can I use the Gram-Schmidt procedure to generate general solutions to a linear ODE?

Is this possible? The question popped into my head as my ODE instructor was teaching us how to solve second-order linear ODEs. Anyway, he said the following things:

A second-order linear ODE has two linearly independent solutions (i.e. the two solutions are not scalar multiples of each other...he did give the formal definition later on). These solutions form a vector space - a solution space - and any linear combination of these solutions is a solution to the ODE. Moreover, these solutions are basis vectors of the space.

I'll pose my question again, more precisely:

Suppose I am able to find a particular solution $y_1$ to a linear ODE of order $n$ in some interval $I \subset \mathbb{R}$ and I define an inner product on $I$. I apply the Gram-Schmidt procedure to $y_1$ and generate functions $y_2,...,y_n$ such that $y_i$ and $y_j$ are pairwise orthonormal for $1 \leq i, j \leq n$. Clearly they are linearly independent. Do these vectors form a solution space for the given $n$-th order ODE? Generally, is a linear combination of the functions $y_i$ a solution to the ODE?

Does my question even make sense? Please let me know if the question, as it stands, is stated imprecisely, incorrectly, etc. Thanks in advance for the help.

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How does applying Gram-Schmidt to a single solution get you $n-1$ other solutions? –  Qiaochu Yuan Sep 28 '12 at 1:45
With $y_1$ and $y_2$, you can generate an orthonormal basis for the space spanned by these functions. I don't know how you would do this with only 1 of them. –  Daryl Sep 28 '12 at 1:46
My mistake, I haven't touched Gram-Schmidt in awhile. Yes, there should be two functions. And the other n-2 solutions would, conceivably, come from applying GSP repeatedly. Are there any other edits I should make to the OP? –  Bachmaninoff Sep 28 '12 at 2:38
hmm...it seems I need to do some more thinking here. It's been TOO long since I touched GSP. Sorry! I see what you're saying, Qiachou. –  Bachmaninoff Sep 28 '12 at 2:44
OP=Owner of Post, so the OP is actually you. –  leo Sep 28 '12 at 3:49