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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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    @QiaochuYuan Thanks.2012-10-05

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