Any eigenvalues of Equation d y 2 x 2 + λ y = 0 \dfrac{dy^2}{x^2}+λy=0 x 2 d y 2 + λ y = 0 must be positive. If y y y satisfies the equation with λ > 0 , \lambda>0, λ > 0 , then
y = c 1 cos λ x + c 2 sin λ x y=c_1\cos\sqrt{\lambda}x+c_2\sin\sqrt{\lambda}x y = c 1 cos λ x + c 2 sin λ x where c 1 c_1 c 1 and c 2 c_2 c 2 are constants. The boundary condition y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 implies that c 1 = 0. c_1=0. c 1 = 0. Therefore
y = c 2 sin λ x y=c_2\sin\sqrt{\lambda}x y = c 2 sin λ x The boundary condition y ( π ) = 0 y(\pi) = 0 y ( π ) = 0 implies that
c 2 sin λ π = 0 c_2\sin\sqrt{\lambda}\pi=0 c 2 sin λ π = 0 To make c 2 sin λ π = 0 c_2\sin\sqrt{\lambda}\pi=0 c 2 sin λ π = 0 with c 2 ≠ 0 , c_2 \not= 0, c 2 = 0 , we must choose λ = n , \sqrt{\lambda} = n, λ = n , where n n n is a positive integer.
Therefore λ n 2 = n 2 \lambda_n^2=n^2 λ n 2 = n 2 is an eigenvalue and
y n = sin ( n x ) y_n=\sin(nx) y n = sin ( n x ) is an associated eigenfunction.
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