Eigenfunctions from eigenvalues unsure

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Homework Help Overview

The discussion revolves around finding eigenvalues and eigenfunctions for the differential equation X''(x) + λX(x) = 0, specifically for the cases where λ = 0, λ = -k², and λ = k², with boundary conditions X(0) = 0 and X'(1) + X(1) = 0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of different values of λ on the solutions, noting that for λ = 0 and λ = -k², no eigenvalues are found. For λ = k², they discuss the resulting equation kcos(k) + sin(k) = 0 and the confusion surrounding the approximation of k values.

Discussion Status

Some participants have suggested that the eigenvalues can be approximated by graphing the functions involved, while others clarify that the approximation holds primarily for small eigenvalues. There is an ongoing exploration of the implications of these approximations and their accuracy as eigenvalues increase.

Contextual Notes

Participants question the reasoning behind focusing on smaller eigenvalues and express uncertainty regarding the professor's guidance on this matter.

WtKemper
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Homework Statement


using X''(x)+ lambda*X(x)=0 find the eigenvalues and eigenfunctions accordingly.
Use the case lambda=0, lambda=-k2, lambda=k2
where k>0

Homework Equations


X(0)=0, X'(1)+X(1)=0


The Attempt at a Solution


I know that for lambda=0
X(x)=C1x+C2
which applying the conditions gives no E.V.

also for lambda=-k2
X(x)=C1cosh(kx)+C2sinh(kx)
and applying the conditions gives no E.V.

for the final case lambda=k2
X(x)=C1cos(kx)+C2sin(kx)

using X(0)=C1=0

X'(x)=C2kcos(kx)

applying second condition then
C2(kcos(k)+sin(k))=0 so if we make the assumption that C2 is not 0 then kcos(k)+sin(k)=0

I've tried multiple things and finally came to dividing by cos(k) so that it becomes
k+tan(k)=0 or k=-tan(k)

but, this is where I get confused. My professor offered the hint that k becomes an approximation so I plotted x and -tan(x) and found where they intersect. This gives a few values but I don't understand how to get a value for k. Normally k=n*pi or some sort of thing. So, my question is how do I use this information to find lambda and the Eigenfunctions for this problem. Any help is appreciated.
 
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It doesn't "give" general values for k. As your professor told you, the best you can do is approximate them, perhaps by graphing as you did. Obviously k= 0 is one value but that is the only one that can be written in a simple way.
 
I figured out what was meant by approximation. As if you plot y=x and y=-tan(x) the points at which they cross close in on the values of pi*n/2 at odd n values. so k~(2n-1)*pi/2.
which makes the Eigenfunction sin(kx)=sin[((2n-1)*pi*x)/2]
 
NO, it does not! All you can say is that the correct eigenvalues are close to (2n-1)pi/2. And, by the way, that is only true for small eigenvalues. As the eigenvalues get larger, that approximation becomes very poor.
 
Sorry I didn't mean to say that it closes in but that it is close for (as you stated) small values. But, for some reason the professor didn't tell us that this was the case and I'm unsure why they are only concerned with the smaller values Thank You for the input and help!
 

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