In this question Understanding solution of PDE using method: separation of variables. (in which I put a bounty..) how and why the eigenfunctions were computed like that?
The origin of the $T_n$ eigenfunctions it's clear, but I don't see the origin of $X_n$. Specifically this is the part of the solution were I got stuck :
Since the matrix has zero determinant, the two linear equations it produces give the same relationship between $A$ and $B$. Thus, we can solve only one of them to determine this relationship. Somewhat arbitrarily, I will choose the second equation, so
$$ B = \frac{\sin\omega L+\sinh\omega L}{\cos\omega L-\cosh\omega L}A = (\sin\omega L+\sinh\omega L)A^* $$
making the obvious substitution for $A^*$. Then, after some manipulations and substitutions, we get
$$ X_n = A_n^* \left[(\cos r_n-\cosh r_n)\left(\cos\frac{r_n x}{L}-\cosh\frac{r_n x}{L}\right) \\ \qquad \quad + (\sin r_n + \sinh r_n)\left( \sin\frac{r_n x}{L}-\sinh\frac{r_n x}{L} \right) \right] $$
Could someone explain please?
It's just that part, I already understand the rest of the long solution.
What part of that quoted answer did you have trouble with? Since the equations determining $A,B$ are homogeneous
\begin{align} \big[\cos(\omega L) - \cosh(\omega L)\big]A + \big[(\sin(\omega L) - \sinh(\omega L)\big]B &= 0 \\ -\big[\sin(\omega L) + \sinh(\omega L)\big]A + \big[\cos(\omega L) - \cosh(\omega L)\big]B &= 0 \end{align}
they both depend on the same constant, i.e, the ratio $\frac{A}{B}$ is a fixed value, dependent only on the eigenvalue $\omega$. So we can set
\begin{align} A &= ~~~c \big[\sin(\omega L) - \sinh(\omega L)\big] \\ B &= -c \big[\cos(\omega L) - \cosh(\omega L)\big] \end{align}
based on the coefficients of the first equation. They satisfy the second equation as well, since
$$ -\big[\sin(\omega L) + \sinh(\omega L)\big]A + \big[\cos(\omega L) - \cosh(\omega L)\big]B \\ = -c\Big[\big[\sin(\omega L) - \sinh(\omega L)\big]\big[\sin(\omega L) +\sinh(\omega L)\big] + \big[\cos(\omega L) - \cosh(\omega L)\big]^2\Big] = 0 $$
Note that the choice of $A,B$ is arbitrary, as the two equations (with the right values of $\omega$) are in fact the same. There are many ways of picking them, and in the end you still have one free constant.
With that result, the final eigenfunction (up to constant) is
$$ X_n(x) = c_n\Big[\big[\sin(\omega_n L) - \sinh(\omega_n L)\big]\big[\cos(\omega_n x) - \cosh(\omega_n x)\big] \\ - \big[\cos(\omega_n L) - \cosh(\omega_n L)\big]\big[\sin(\omega_n x) - \sinh(\omega_n x)\big]\Big] $$