prove $\left\{\begin{array}{l}X^{\prime \prime}+\lambda X=0,00$ without solving Can you help me some idea to prove it or some similar question about it. I know prove $\lambda>0$ is easy when solving it, it's just like normal differential equation.

The idea is that the differential equation either has sines and cosines for solutions (if $\lambda > 0$) or exponentials (if $\lambda < 0$) or linear ($\lambda = 0$).
The second boundary condition looks a bit strange.  If it had been $X(1) = 0$, that would have been enough to conclude it must be trig functions right there.
The $\lambda = 0$ case can be ruled out because $X(1) + X'(1) = 0$ means that there must be a positive slope to a negative value, or vice versa, which doesn't work starting from the origin except with a zero solution.
