Consider the Sturm - Liouville problem
$$(1+x^4)y''(x)+4x^3y'(x)+\lambda y(x)=0,~0<x<\pi,$$ equipped with the boundary conditions $y'(0)=y(\pi)=0.$
$(i)$ Prove that the eigenvalues $\lambda_n,~n\geq 1$ of the above problem are all positive.
$(ii)$ If $y_n,~y_m$ are eigenfunctions corresponding to different eigenvalues $\lambda_n~\lambda_m$, respectively, prove that: $$\int_{0}^{\pi}(1+x^4)y'_n(x)y'_m(x)dx=0.$$
Some thoughts. The problem does not have constant coefficients, making it impossible to work with the characteristic polynomial, neither is of a special form, for example Euler, in order to end up to a constant- coefficient-problem with the correct substitution. Am I missing this "correct substitution" or is there something else, in order to find the eigenvalues - eigenvectors?
Thanks a lot in advance.
You don't need to find the solutions to the ODE to solve this problem. All the information you need to use is the structure of the ODE itself. Here are some pointers on how to proceed:
(i) Multiply the ODE by $y$ to get
$$[(1+x^4)y']'y + \lambda y^2 = 0$$
Integrate this over $[0,\pi]$ to get
$$\int_0^\pi [(1+x^4)y'(x)]'y(x)\,{\rm d}x + \lambda \int_0^\pi y^2(x)\,{\rm d}x = 0$$
Apply integration by parts on the first term and use the boundary condition to simplify and show that this gives an equation on the form $\lambda\cdot {\rm positive} = {\rm positive}$.
(ii) Use integration by parts
$$\int_0^\pi (1+x^4)y_n'y_m'\,{\rm d}x = [(1+x^4)y_n'y_m]_0^\pi - \int_0^\pi[ (1+x^4)y_n']'y_m\,{\rm d}x$$ Use the ODE $[(1+x^4)y_n']' = -\lambda_n y_n$ and the boundary conditions to simplify the equation above. Finally note that the left hand side of the equation above is symmetric in $n$ and $m$ so the result we found above also applies if we interchange $n$ and $m$. Use this to conclude that $\int_0^\pi y_ny_m{\rm d}x = 0$ when $\lambda_n \not=\lambda_m$.