Sturm-Liouville eigenvalues

[renegade05]

I am tackling this S-L problem :

\(\displaystyle -(xX')' = \frac{\lambda}{x} X \quad 1<x<e\)

\(\displaystyle X'(1) - X(1) = 0 \quad X'(e)+10X(e)=0\)

Fast forward some work showing that [FONT=MathJax_Math-italic]λ [/FONT]has to be positive.

I get the following result:

\(\displaystyle X(x) = Acos(\sqrt{\lambda}\ln{x}) + Bsin(\sqrt{\lambda} \ln{x})\)
\(\displaystyle X'(x) = -Asin(\sqrt{\lambda}\ln{x}) \frac{\sqrt{\lambda}}{x}+ Bcos(\sqrt{\lambda} \ln{x})\frac{\sqrt{\lambda}}{x}\)

\(\displaystyle X'(1) - X(1) = B\sqrt{\lambda} - A = 0 \rightarrow B\sqrt{\lambda} = A \)
\(\displaystyle X'(e) + 10 X(e) = -Asin(\sqrt{\lambda}) \frac{\sqrt{\lambda}}{e}+ Acos(\sqrt{\lambda})\frac{1}{e} + 10Acos(\sqrt{\lambda}) + 10Asin(\sqrt{\lambda})\frac{1}{\sqrt{\lambda}}=0\)
\(\displaystyle \rightarrow Acos(\sqrt{\lambda})\frac{1}{e} + 10Acos(\sqrt{\lambda}) =Asin(\sqrt{\lambda}) \frac{\sqrt{\lambda}}{e} - 10Asin(\sqrt{\lambda})\frac{1}{\sqrt{\lambda}}\)
\(\displaystyle \rightarrow (\frac{1}{e} + 10)cos(\sqrt{\lambda}) =(\frac{\sqrt{\lambda}}{e} - \frac{10}{\sqrt{\lambda}})sin(\sqrt{\lambda}) \)
\(\displaystyle \rightarrow \tan{\sqrt{\lambda}}=\frac{\sqrt{\lambda}(10e+1)}{\lambda-10e}\)

I think all this algebra is right....

But I am stuck trying to solve the eigenvalues now... Please help