1. Identify p(x), q(x), w(x) alpha, beta, gamma, delta, solve for the eigenvalues and eigenfunctions, and work out the eigenfunction expansion of the given function f. If the characteristic equation is too difficult to solve analytically, state that and proceed with the rest of the problem as though the lambda_n's were known.
(b) y" + lambda y = 0,
y'(0) = 0, y(L) = 0, f(x) = 1.
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We have y''+ lambda*y=0 with y'(0)=0 and y(L)=0, so:
If lambda=0 --> y''=0 --> y(x)=Ax+B, y'(0)=0 ---> A=0 and y(L)=0 ---> B=0. Therefore, lambda=0 gives no meaningful eigenvectors.
y(x)= A*cos(sqrt(lambda)*x)+ B*sin(sqrt(lambda)*x)
y'(x)= -sqrt(lambda)*A*sin(sqrt(lambda)*x)+ sqrt(lambda)*B*cos(sqrt(lambda)*x)
so: y'(0)= B*sqrt(lambda)=0, so B=0. We also have that:
This solution shows how to solve for eigenvalues and eigenfunctions and work out an eigenfunction expansion.