How does one find the eigenvalues and eigenfunctions of

-(x^2y')'=lambda *y

with 1<x<e, and y(1)=y(e)=0

I'm completely lost. Thank you!

Printable View

- May 8th 2011, 05:48 PMmorganforSturm-Liouville
How does one find the eigenvalues and eigenfunctions of

-(x^2y')'=lambda *y

with 1<x<e, and y(1)=y(e)=0

I'm completely lost. Thank you! - May 8th 2011, 09:14 PMTheEmptySet
This equation is of type Cauchy-Euler

Using the anzant

gives the characteristic equation

From the boundary conditions we can see we need periodic solutions so the roots of the characteristic equation must be complex. Solving the above for r gives

This implies that lambda must be positive

This gives solutions

Using the boundary conditions to solve for lambda gives

This gives

Now with

This gives

- May 8th 2011, 10:12 PMmorganfor
thanks!!