I want to gain more intuition on the spectral theorem, so I came up with an exercise to test it in a concrete case. I thought considering the second order ODE: $$A\frac{d^2}{dx^2}f(x)+Bf(x)=\lambda f(x)$$ By showing that the linear operator: $$Df=A\frac{d^2}{dx^2}f(x)+Bf(x)$$ Satisfies an equation: $$Df=\lambda f$$ Where $\lambda$ is an eigenvalue of $L$ in a suitable function space $F$.
I know that this DE has solutions, since it is only a reformulation of the more known equation for the harmonic simple motion: $$A\frac{d^2}{dx^2}f(x)+(B-\lambda)f(x)=0$$ It is known that the solution to this equation has the general form: $$S\sin (x)+T\cos(x)$$ and these trigonometric functions are know to form a basis for the $L^2([-\pi,\pi])$ space of functions; so, if I extract the matrix of the operator, I get: $$D=\begin{bmatrix}A\frac{d^2}{dx^2}&0\\0&B\end{bmatrix}$$ The operator is normal, because it satisfies: $$DD^*=D^*D$$ And from this we can conclude that the eigenvalues of $D$ should be the roots of the polynomial: $$z^2-(A\frac{d^2}{dx^2}+B)z+AB\frac{d^2}{dx^2}$$ Which, by the spectral theorem, are real numbers, and should be: $$\begin{array}&\lambda_0=A\frac{d^2}{dx^2}\\\lambda_1=B\end{array}$$ But here I start entering muddy waters. I've never seen or done the rest of the process, and don't really know where to go from here to "solve" the equation. More concretely, suppose the boundary conditions are given so that a solution of the form: $$f(x)=S\sin(x)+T\cos(x)$$ Is required (which we don't yet know). How can I get there from the data I collected? I.e., how can I get the eigenvectors $f_0$ and $f_1$ related to $\lambda_0$ and $\lambda_1$, respectively? Thanks in advance.