- #31
AntSC
- 65
- 3
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?
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?
