Applying Boundary Conditions in Homogeneous ODE Problems

[AntSC]

Fantastic. Didn't see that at all. Thanks very much!

For $\lambda=1 \Rightarrow u(x)=e^{-x}[Ax+B]$
So $u(0)=0 \Rightarrow B=0$ and $u(1)=0 \Rightarrow A=0$ and therefore $u(x)=0$

For $\lambda>1 \Rightarrow u(x)=e^{-x}[A\cos (\sqrt{\lambda-1})x+B\sin (\sqrt{\lambda-1})x]$
So $u(0)=0 \Rightarrow A=0$ and $u(1)=0 \Rightarrow$ either $B=0$ or $\sin (\sqrt{\lambda-1})=0 \Rightarrow (\sqrt{\lambda-1})=n\pi$

So the eigenvalues are $\lambda =1+\pi^{2}n^{2}$
And the eigenfunctions are $u(x)=Be^{-x}\sin (n\pi x)$

All correct?

 

[ehild]

Yes, it looks correct

ehild

 

[AntSC]

Great. Thanks so much for all your help!
I'm pleased to put this one to bed and come out having cleared some things up.
Cheers!