First separate variables. Let: $\Big {[}u(x,t) = X(x)T(t)\Big {]}$ in $\Big {[}u_{tt}(x,t) + K u_{xxxx} = 0, K > 0\Big {]}$
Then:$$\Big {[}u_{tt}(x,t) = X(x)T(t), u_{xxxx}(x,t) = X(x)T(t)\Big {]}$$
$$\Big {[}u_{tt}(x,t) + K u_{xxxx} = X(x)T(t) + K X(x)T(t) = 0\Big {]}$$
$$\Big {[}\frac{X(x)}{X(x)} = \frac{-T(t)}{K T(t)} = \lambda\Big {]}$$
For some constant, as both sides are independant of the other respective variable. Now, we have by assumptions that $\Big {[}\lambda = c^4 > 0, K = k^2 > 0\Big {]}$
So we are left with two ODEs in the form: $$\Big {[}X(x) = c^4 X(x), T(t) = -c^4 k^2 T(t)\Big {]}$$
Which yield solutions: $$\Big {[}X(x) = A \cosh(c x) + B \sinh(c x) + C \cos(c x) + D \sin(c x)\Big {]}$$
$$\Big {[}T(t) = A \cos(c^2 k t) + B \sin(c^2 k t)\Big {]}$$
Using the two boundary conditions: $\Big {[}u(0,t) = 0 = u_x(0,t)\Big {]}$
$$\Big {[}X(0) = A \cosh(c 0) + B \sinh(c 0) + C \cos(c 0) + D \sin(c 0) = A + C = 0, A = -C\Big {]}$$
$$\Big {[}X(0) = A c \sinh(c 0) + B c \cosh(c 0) - C c \sin(c 0) + D c \cos(c 0) = B c + D c, D = -B\Big {]}$$
As we disregard the case where: $$\Big {[}X(0) \neq 0, X(x) \neq 0 \implies T(t)=0\Big {]}$$
We plug into the 3rd and 4th boundary conditions.
$$\Big {[}u_{xx}(l,t) = 0 = u_{xxx}(l,t)\Big {]}$$
$$X(l) = A c^2 \cosh(c l) + B c^2 \sinh(c l) - C c^2 \cos(c l) - D c^2 \sin(c l)$$
$$=A c^2 \cosh(c l) + B c^2 \sinh(c l) + A c^2 \cos(c l) + B c^2 \sin(c l) $$$$= A c^2 (\cosh(c l) + \cos(c l)) + B c^2 (\sinh(c l) + \sin(c l)) = 0$$
$$\Big {[}X(l) = A c^3 (\sinh(c l) - \sin(c l)) + B c^3 (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
Yielding:
$$\Big {[}A (\cosh(c l) + \cos(c l)) + B (\sinh(c l) + \sin(c l)) = 0\Big {]}$$
$$\Big {[}A (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}A = -B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))}\Big {]}$$
$$\Big {[}-B \frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l)) + B (\cosh(c l) + \cos(c l)) = 0\Big {]}$$
$$\Big {[}-\frac{(\sinh(c l) + \sin(c l))}{(\cosh(c l) + \cos(c l))} (\sinh(c l) - \sin(c l) + (\cosh(c l) + \cos(c l)))\Big {]}$$
$$\Big {[}-\sinh(c l)^2 + \sin(c l)^2 + \cosh(c l)^2 + 2 \cosh(c l) \cos(c l) + \cos(c l)^2 =0\Big {]}$$
Where we used the facts that $$\Big {[}B = 0 \implies X(x) = 0, (\cosh(c l) + \cos(c l)) = 0 \implies B = A = 0\Big {]}$$
Which we disregard as were not so interested in the trivial case.
$$\Big {[} \cosh(c l)^2 - \sinh(c l)^2 + \sin(c l)^2+ \cos(c l)^2 + 2 \cosh(c l) \cos(c l) = 1 + 1 + 2 \cosh(c l) \cos(c l) = 0\Big {]}$$
As:$ \Big {[}\cosh^2 - \sinh^2 = 1, \cos^2 + \sin^2 = 1\Big {]}$
So our eigenvalues are those c which satisfy$ \Big {[}\cosh(c l) \cos(c l) = -1 \blacksquare$
Next consider: $$\Big {[}\int_0^l X_n(x)X_m(x) dx, n \neq m, \lambda_n \neq \lambda_m\Big {]}$$
Then $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = \int_0^l \lambda_n X_n(x)X_m(x) - \lambda_m X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}\int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} - \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx\Big {]}$$
$$\Big {[}= 0 - (X_n(x)X_m(x) - X_n(x)X_m(x))_{x=(0,l)} + \int_0^l X_n(x)X_m(x) - X_n(x)X_m(x) dx = 0 + 0 = 0\Big {]}$$
Using integration by parts, and the fact that our boundary conditions vanish at:
$$\Big {[}X(0) = 0, X(0) = 0, X(l) = 0, X(l) = 0\Big {]}$$
Then we have $$\Big {[}(\lambda_n - \lambda_m) \int_0^l X_n(x)X_m(x) dx = 0, \lambda_n \neq \lambda_m \neq 0 \implies \int_0^l X_n(x)X_m(x) dx = 0\Big {]}$$
And our eigenfunctions are orthogonal.
Let our differential operator be$$ \Big {[}\mathcal{I}, st. \mathcal{I} X = \lambda X, \mathcal{I} Y = \lambda Y\Big {]}$$
We show that: $$\Big {[}\langle \mathcal{I}X,Y\rangle = \langle X,\mathcal{I}^*Y\rangle\Big {]}$$
$$\Big {[}\langle \mathcal{I}X,Y\rangle = \int_0^l \mathcal{I}X(x)Y(x)^* dx = \int_0^l \lambda X(x)Y(x)^* dx = \int_0^l X(x)\lambda^* Y(x)^* dx = \int_0^l X(x)\mathcal{I}^*Y(x)^* dx = \langle X,\mathcal{I}^*Y\rangle\Big {]}$$
As we had by assumption our eigenvalues were real. Then our operator is hermetian, so the eigenfuncitons of different eigenvalues are linearly independant, and the eigenfunctions of the same eigenvalue are linearly dependant.
