A problem in using boundary condition in differential equation

Question

The general solution of the heat equation is $$X(x) = a \sin \omega x + b \cos \omega x$$

For the boundary conditions of $$X(0) = 0\\ X(L) = 0$$ we will have $b=0$ and $a \sin \omega L$. Therefore the solution is $$u_n(x,t)=e^{-n^2\pi^2kt/L^2} \sin(n\pi x/L)$$

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Although it is unnecessary, but it is not incorrect to interpret the above boundary condisions as $$X'(0) = 0\\ X'(L) = 0$$ By applying new boundary conditions in $$X'(L) = \omega (-a \cos \omega L + b \sin \omega L)$$ we will have $$X(x) = a \cos(n \pi x/L)$$ and then the general solution leads to $$u_n(x,t)=e^{-n^2\pi^2kt/L^2} \cos(n\pi x/L)$$
What is the reason for this difference?

The reason that I am asking this question is that the first approach is the solution for the Dirichlet boundaries and the second for the Neumann boundaries. If we have a mix of Dirichlet-Neumann boundaries, which approach should be used? Shouldn't we get the same result by both approaches?

Answer 1

Citation :

Although it is unnecessary, but it is not incorrect to interpret the above boundary condisions as $$X'(0) = 0\\ X'(L) = 0$$

No, it is incorrect.

$\begin{cases}X(0)=0\\X(L)=0\end{cases}\quad$ are different boundary conditions from $\quad\begin{cases}X'(0)=0\\X'(L)=0\end{cases}\quad$

With different boundary conditions, it is not surprising to obtain different solutions.