To put it in simple algebraic terms: your differential operator

e^-x  d/dx (  e^x f' [x]   == λ e^x  Y[ x ] 

is not symmetric with respect to the Lebesgue measure dx.

Write the Hilbert space product on (0,Pi) with functions f,g with f having zeros at 0, Pi

int  [f [x]^*   e^-x  d/dx ( e^x g ' [x]  ) dx ==
- int  d/dx ( e^-x f [x]^*)   e^x  g' [x]  dx  ==
 Int  d/dx ( e^x[  d/dx (e^-x f [x]^* ) )     g[x]  dx 

That means, that without a suitable measure of integration, this is a pair of different adjoint differential operators

(  e^-x  d/dx ( e^x d/dx #   & )^* ==   ( d/dx [ e^x  d/dx ( e^-x)  #  ]&)

To make it symmetric with respect to a double integration by parts, you simply have to insert the measure of integration that cancels the e^-x in front of D

int  f [x]^*   e^-x  d/dx ( e^x g ' [x]  ) ( e^x dx)   ==
-   int e^x  f'  [x]^*    g ' [x]  (  dx)   == 
int  e^-x d/dx( e^x f  [x]^* )    g ' [x]  ) ( e^x dx) 

It follows that D is symmetric in the Hilbertspace L_2 ( (0,Pi) , e^x dx ) with boundary value 0. it has a complete set of real eigenvalues and eigenfunctions with functions of different eigenvalues being orthogonal.

Even in this setting the operator is not essential selfadjoint, because the domains of D and D* are different, the dual space of adjoints g(x)^* has no contraint for zeros at (0, Pi).The different selfdjoint extensions belong to the known families of basis functions with any combibations of boundary conditions at 0, Pi

This fact is the main difference of linear operators in finite and infinite dimensional linear spaces.

Regards Roland