MHB Find Eigenvalues for $$y''+\lambda y=0$$

[mathmari]

That sounds very plausible. (Mmm)

Actually, $y=0$ is always a solution, regardless of the value of $\lambda$, so $\lambda=0$ is not really the corresponding eigenvalue.
So it makes sense to leave that one out.

Furthermore, the eigenfunctions for $-\ln^2(2\pm\sqrt 3)$ are identical.
That is because $$\frac 1{2-\sqrt 3} = 2+\sqrt 3$$.
So we might consider them as the same.

Great! (Whew)

Have you tried to calculate the Wronskian for 2 eigenfunctions of which 1 has an imaginary eigenvalue? (Wondering)

So should I try it for $u(x)=\sin{(x(-i \ln{(2+\sqrt{3})}+2 \pi n))}$ and $v(x)=v^*(x)=(2+\sqrt{3})^x-(2+\sqrt{3})^{-x}$ ?

 

[I like Serena]

So should I try it for $u(x)=\sin{(x(-i \ln{(2+\sqrt{3})}+2 \pi n))}$ and $v(x)=v^*(x)=(2+\sqrt{3})^x-(2+\sqrt{3})^{-x}$ ?

Yes... (Sweating)

 

[mathmari]

Yes... (Sweating)

If I have done right the calculations, the Wronskian at $1$ is not equal to $0$! (Wink) (Party) (Dance)

 

[mathmari]

Thank you for your help! (Bow)