Finding Eigen Values/functions

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SUMMARY

The discussion focuses on finding the eigenvalues and eigenfunctions for the differential equation y'' + λy = 0 with boundary conditions y(0) + y'(0) = 0 and y(π) = 0. The general solution is expressed as y(x) = A sin(x√λ) + B cos(x√λ). The conclusion drawn is that for the boundary condition y(π) = 0 to hold, B must equal 0 and √λ must be a positive integer n. However, a counterexample is provided, indicating that λ does not necessarily have to be an integer, challenging the initial assumption.

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Homework Statement


I have to find the eigenvalues&function of the eqn:
[tex]y''+\lambda y =0[/tex]
With the boundary conditions:
[tex]y(0)+y'(0) = 0[/tex] and [tex]y(\pi) =0[/tex]

Homework Equations


The Attempt at a Solution


I get the general equations, okay, but am having trouble due to the boundary conditions.
Assuming [tex]\lambda>0[/tex], then get the general solution:
[tex]y(x)=A\sin(x\sqrt{\lambda})+B\cos(x\sqrt{\lambda})[/tex]

The best i can do now is that:
[tex]y(\pi)=A\sin(\sqrt{\lambda}\pi)+B\cos(\sqrt{\lambda}\pi)=0[/tex]
Which can only be valid if B = 0 and [tex]\sqrt{\lambda} = n[/tex] where n is a positive integer.
Now I run into problems, the second bc gives:
[tex]y(0)=A\sin(n 0)=0[/tex]
[tex]y'(0)=nA\cos( n 0)=nA[/tex]
So:
[tex]y(0)+y'(0) = 0 + nA = 0[/tex]

Am I on the right track, or can anyone see where I am messing up?
Thanks.
 
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ShowerHead said:
[tex]y(\pi)=A\sin(\sqrt{\lambda}\pi)+B\cos(\sqrt{\lambda}\pi)=0[/tex]
Which can only be valid if B = 0 and [tex]\sqrt{\lambda} = n[/tex] where n is a positive integer.

No, as a simple counterexample consider [itex]\lambda=\frac{1}{16}[/itex] and [itex]A=-B[/itex]. If [itex]\sqrt{\lambda}[/itex] had to be an integer, then [itex]B[/itex] would have to be zero; but there's no reason to assume [itex]\sqrt{\lambda}[/itex] must be an integer.
 

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