For the boundary value problem
$$y''+\lambda y=0,\ \ y(0)-y'(0)=0,\ \ y(1)=0$$
I believe only the trivial solution applies when $\lambda =0$. When $\lambda <0$ I get solutions of the form $c_1e^{-\sqrt{-\lambda}t}+c_2e^{\sqrt{-\lambda}t}$ and applying the boundary conditions,
$$c_1(1+\sqrt{-\lambda})+c_2(1-\sqrt{-\lambda})=0$$
$$c_1e^{-\sqrt{-\lambda}}+c_2e^{\sqrt{-\lambda}}=0$$
The second implies
$$c_1=-c_2e^{2\sqrt{-\lambda}}$$
which we can substitute into the first equation to get
$$1-\sqrt{-\lambda}=(1+\sqrt{-\lambda})e^{2\sqrt{-\lambda}}$$
I see no clear path to a solution for $\lambda$ but I do remark that the right-hand side is positive hence we require
$$1-\sqrt{-\lambda} > 0$$
so that $0>\lambda > -1$. That all such $\lambda$ would allow for a solution seems unlikely to me, but I can see no further solution method, except perhaps by using technology to plot the functions of $\lambda$ and find intersections. Is there a more satisfying solution?