y''+λy = 0 ... y'(0)=0 ... y'(π)=0?

Question

the problem says that the eigenvalues are all nonnegative, so you don't need to check that, simply what happens when λ=0 and when λ is positive.

Then says, determine if λ=0 is an eigenvalue, then find the positive eignvalues and associated eigenfunctions.

I've tried this several times, but can't quite figure out what I'm doing wrong.

btw the answer in the book says:

Eigenvalue λ[subscript 0]=0 with eigenfunction y[subscript ](x) ≡ 1, and positive eigenvalues {n^2} with associated eigenfunctions {cos nx} for n=1,2,3...

Answer 1

If \lambda = 0 then your differential equation is

y'' = 0
implying that
y(x) = A/2*x^2 + B*x + C
y'(x) = A*x + B
y'(0) = 0 so B=0
y'(\pi) = A*pi = 0 so A=0

so if \lambda = 0, then y(x) = c (a constant) but you can't solve for it. If you use normalization, then y(x) = 1.

If \lamba>0, then you have a harmonic oscillator differential equation. Your solutions will only be homogeneous and of the form

y(x) = A*cos(\sqrt(\lambda)x) + B*sin(\sqrt(\lambda)x)

if you let \lambda = n^2 then you get

y(x) = A*cos(nx) + B*sin(nx)
Applying BC's

y'(x) = -nA*sin(nx) + nB*cos(nx)
y'(0) = nB = 0 so B=0
y'(pi) = -nA*sin(npi) = 0 because sin(n*pi) = 0 for n=1,2,3...etc So your eigenvalues are \lamba = 1,4,9,....

so your eigenfunctions are y(x) = A*cos(nx), you can solve for the coefficients if you choose to normalize.

Answer 2

If k ( instead of lambda) is 0
y´´=0 so y´=m and y=mx+n so m must be 0 and y=n (any constant)
If k is positive call it w^2
The solution is
y=Acos(wx-phi)
y´=-Aw sin(wx-phi)
y´(0) =0 so sin (-phi)=0 and phi =0
y´(pi))=-Aw sin(w*pi)=0 so wpi=n*pi so w= n( an integer)