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.

We compute the Rayleigh quotient. Multiply your ODE by $X$ (assuming $X \ne 0$ is non-trivial) and integrate over the domain $x \in [0, 1]$
\begin{align}
- \lambda X^{2} &= X X'' \\
\implies - \lambda \int_{0}^{1} X^{2} dx &= \int_{0}^{1} X X'' dx \\
&= XX' \bigg \lvert_{0}^{1} - \int_{0}^{1} X'^{2} dx \\
&= - X(1)^{2} - \int_{0}^{1} X'^{2} dx
\end{align}
where we have used integration by parts and the boundary conditions to get the last line. Hence
$$\lambda = \frac{X(1)^{2} + \int_{0}^{1} X'^{2} dx}{\int_{0}^{1} X^{2} dx} > 0$$
as the integral of a non-negative function is non-negative and the numerator is always positive.
