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    $\begingroup$ "but it also claims zero is an eigenvalue" No, it doesn't. It just claims when $\lambda=0$, the BVP has a solution. If one wants to find the eigenvalue, DEigenvalues in principle can be used, but something like DEigenvalues[{-y''[x] + NeumannValue[-y[x], x == 1] + NeumannValue[y[x], x == 0]}, y[x], {x, 0, 1}, 5] doesn't work, NDEigenvalues works, though… $\endgroup$
    – xzczd
    Commented Feb 2, 2018 at 11:39
  • $\begingroup$ @xzczd Ok, thanks. I am used to using DSolve as above to find the eigenvalues since it is easier to use than the syntax of DEigenvalues, but I did not realize that the trivial solution will be also given by DSolve. This was the confusing part for me. So I should switch to DEigenvalues from now on to check my solution. $\endgroup$
    – Nasser
    Commented Feb 2, 2018 at 11:46
  • $\begingroup$ I agree NeumannValue is somewhat hard to use. (You may want to read this post. ) Hope DEigenvalues etc. will be more flexible and powerful in future versions. $\endgroup$
    – xzczd
    Commented Feb 2, 2018 at 11:56
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    $\begingroup$ A simple reminder that for an eigenvalue you have to have a nonzero eigenvector would have been enough. Just a brain fart. :) $\endgroup$
    – Michael E2
    Commented Feb 2, 2018 at 18:57
  • $\begingroup$ @MichaelE2 The \underline{Summary} makes me believe the details were already written before :) $\endgroup$
    – anderstood
    Commented Feb 2, 2018 at 19:22