= Solution
The equation is $y''+(1+\lambda)y=0$. If $1+\lambda\leq0$, the <linear> or <hyperbolic> branches cannot meet both homogeneous endpoint conditions nontrivially. Otherwise $y=C\sin(kx)$, $k=\sqrt{1+\lambda}$, and $y'(1)=Ck\cos k=0$ gives $k_n=(n+1/2)\pi$. Therefore
$$
\boxed{\lambda_n=(n+\tfrac12)^2\pi^2-1,\qquad\phi_n(x)=\sqrt2\sin[(n+\tfrac12)\pi x],\quad n=0,1,\ldots.}
$$
The factor $\sqrt2$ makes the <eigenfunctions> unit normalized. All the <eigenvalues> are positive, in particular none is zero.
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