I've been on this question for ages now, and eventually I came to this solution, however I'm very unsure on if I'm correct or if my approach is right.
Question:
Let $\hat{H}=\frac{d^2}{dx^2}-2\frac{d}{dx}+1$ be a linear differential operator defined on the space $S=\{f\in C^\infty[0,\pi]|f(0)=f(\pi)=0\}$. Find all real positive eigenvalues of $\hat{H}$ and their corresponding eigenfunctions. Should you expect the eigenfunctions to be orthogonal to each other?
My attempt:
To find the eigenvalues of $\hat{H}$, we want to solve the BVP $\hat{H}f+\lambda f=0$ with boundary conditions $f(0)=0,f(\pi)=0$. This corresponds to a second order linear ODE
$f''-2f'+(\lambda+1)f=0$.
Thus our corresponding characteristic equation is
$r^2-2r+(\lambda+1)=0$,
which we can solve to obtain
$r=1\pm \sqrt{\lambda}i$
and hence our general solution to the ODE is
$f(x)=e^x(A\mathrm{cos}(\sqrt{\lambda}x)+B\mathrm{sin}(\sqrt{\lambda}x))$.
Imposing our first boundary condition shows that $A=0$.
Imposing the second condition leads us to the equation
$e^\pi(B\mathrm{sin}(\sqrt{\lambda}\pi))=0\implies B\mathrm{sin}(\sqrt{\lambda}\pi)=0$.
The only way for this equality to hold true is if $\sqrt{\lambda}\pi$ is identically an integer multiple of $\pi$. Therefore, $\sqrt{\lambda}=n, n\in \mathbb{Z} \implies \lambda = n^2, n\in \mathbb{Z}$
Thus we have that our eigenvalues of $\hat{H}$ are $\lambda=n^2,n\in\mathbb{Z}$ and our corresponding eigenfunctions are $f_n=e^x(B\mathrm{sin}(n\pi))$
As for the orthogonality of the eigenfunctions, our adjoint operator $\hat{H}^+ =\frac{d^2}{dx^2}+2\frac{d}{dx}+1$. Since $\hat{H}\ne \hat{H}^+$, $\hat{H}$ is not self adjoint in $S$ and thus our eigenfunctions may not necessarily be orthogonal to each other.