given the legendre equation $(1-x^2)y'' - 2xy' + by = f(x)$ why can the solution be a series of legendre polynomials $y(x) = \sum_{n=0}^{\infty}a_n P_n(x)$? i thought legndre solves the homogenous version of the equation so wouldn't a series of legendre polynomials all equal 0, or even more isn't there only 1 legendre polynomial a solution to the equation?
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The $n$-th Legendre polynomial solves the homogeneous equation for $b=n(n+1)$. Not every Legendre polynomial solves the homogeneous equation for any $b$. However, since the Legendre polynomials form a basis of $L^2(-1,1 )$, any solution to the equation that lies in that space can be expressed as a series of Legendre polynomials, so the ansatz for $y(x)= \ldots$ is sensible. – DominikS Jan 09 '24 at 09:22
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@DominikS what do u mean by a L2(−1,1) – Beast Jan 09 '24 at 09:29
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The space of all square integrable functions over that interval, see here. – DominikS Jan 09 '24 at 09:30
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The fact that it is a Sturm Liouville diff. equation could help you. – Jean Marie Jan 09 '24 at 10:50