Any eigenvalues of Equation x2dy2+λy=0 must be positive. If y satisfies the equation with λ>0, then
y=c1cosλx+c2sinλxwhere c1 and c2 are constants. The boundary condition y(0)=0 implies that c1=0. Therefore
y=c2sinλxThe boundary condition y(π)=0 implies that
c2sinλπ=0To make c2sinλπ=0 with c2=0, we must choose λ=n, where n is a positive integer.
Therefore λn2=n2 is an eigenvalue and
yn=sin(nx) is an associated eigenfunction.
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