Fourth order PDE solution for vibrating beam rigidly fastened at one end and simply fastened at other end. I have Partial Differential equation in the form:
$$ \frac{\partial^2 y}{\partial t^2} + \frac{\partial^4 y}{\partial x^4} =0, \quad 0<x<1 $$
Vibrating Beam
Boundary Conditions:
$$ y(0,t) = \frac{\partial y}{\partial x}(0,t) = 0 \\ y(1,t) = \frac{\partial^2 y}{\partial x^2}(1,t) = 0 $$
Initial conditions:
$$y(x,0) = 0.1\sin(\pi x), \ \frac{\partial y}{\partial t}(x,0) = 0$$
I have come up to the point where I have solved the boundary conditions and got the four equations:
$A+C=0$
$B+D=0$
$A\sinh(\beta) + B\cosh(\beta) + C\cos(\beta) + D\sin(\beta) = 0 \\$
$A\sinh(\beta)+B\cosh(\beta)-C\sin(\beta)-D\cos(\beta) = 0$
Not sure what to do next. I am supposed to find the Eigenvalues and Eigenfunctions.
 A: We have $C=-A$ and $D=-B$, so
$$ A(\sinh \beta - \cos \beta) + B(\cosh \beta - \sin \beta) = 0 $$
$$ A(\sinh \beta + \sin \beta) + B(\cosh \beta + \cos \beta) = 0 $$
This guarantees a solution if the determinant of the coefficient matrix is zero
$$\left|\begin{matrix} \sinh\beta - \cos\beta && \cosh\beta -\sin\beta \\
\sinh\beta + \sin\beta && \cosh\beta + \cos\beta \end{matrix}\right| = 0 $$ 
which, after some manipulation, gives
$$ e^{-\beta} = \sin\beta - \cos \beta = \sqrt{2}\sin\left(\beta - \frac{\pi}{4}\right) $$
You can solve this graphically as the intersection of the two functions. The first eigenvalue is $\beta_0 \approx 1.038415$. Since $e^{-\beta}$ decays rapidly, the larger solutions are approximately $\beta_n \sim (n+1/4)\pi $

EDIT: I'm not sure if your equations are actually correct. I tried solving for the eigenfunction and got
$$ X(x) = A\cosh \beta x + B\sinh \beta x + C\cos\beta x + D\sin\beta x $$
And the four boundary conditions give
$$ X(0) = A + C = 0 $$
$$ X'(0) = B + D = 0 $$
$$ X(1) = A\cosh\beta + B\sinh\beta + C\cos\beta + D\sin\beta = 0 $$
$$ X''(1) = A\cosh\beta + B\sinh\beta - C\cos\beta - D\sin\beta = 0 $$
In this case, the eigenvalues satisfy
$$\left|\begin{matrix} \cosh\beta - \cos\beta && \sinh\beta -\sin\beta \\
\cosh\beta + \cos\beta && \sinh\beta + \sin\beta \end{matrix}\right| = 0 $$
or
$$ \tanh \beta = \tan\beta $$
Since $\tanh\beta$ very quickly goes to $1$, $\beta_n \sim (n + 1/4)\pi, \ n \ge 1$. This is a very good approximation; the first zero is $\beta_1 \approx 3.9266 = 5\pi/4 - 0.0004 $

EDIT 2: For the eigenfunctions, we have
$$ X(x) = A(\cosh\beta x - \cos\beta x) + B(\sinh\beta x - \sin\beta x) $$
where $(A,B)$ satisfy
$$ A(\cosh\beta-\cos\beta) + B(\sinh\beta-\sin\beta) = 0 $$
$$ A(\cosh\beta+\cos\beta) + B(\sinh\beta+\sin\beta) = 0 $$
Note that these are actually the same equation (since the determinant is zero), so we can just pick any arbitrary pair that satisfies one of the above. For example, let $A = \sinh\beta-\sin\beta$ and $-B = \cosh\beta-\cos\beta$, then the eigenfunction is (up to a constant)
$$ X_n(x) = (\sinh\beta_n-\sin\beta_n)(\cosh\beta_n x - \cos\beta_n x) - (\cosh\beta_n-\cos\beta_n)(\sinh\beta_n x - \sin\beta_n x) $$
To show these functions are mutually orthogonal, we use the definition of orthogonality
$$ \int_0^1 X_n(x) X_m(x)\ dx = 0, \ m \ne n $$
The integral is a bit messy, but it should reduce down to some expression in terms of the determinant.
EDIT 3: The B.C.s are $X(0) = X(1) = X'(0) = X''(1) = 0$. Using integration by parts, we can show that
\begin{align} 
\int_0^1 {X_m}^{(4)} X_n &= {X_m}'''X_n\Bigg|_0^1 - \int_0^1 {X_m}'''{X_n}', && X_n(0)=X_n(1)=0 \\
&= -{X_m}''{X_n}'\Bigg|_0^1 + \int_0^1 {X_m}''{X_n}'', && {X_m}''(1) = {X_n}'(0) = 0 \\
&= {X_m}'{X_n}''\Bigg|_0^1 - \int_0^1 {X_m}'{X_n}''', && {X_m}'(0) = {X_n}''(1) = 0 \\ 
&= -X_m{X_n}^{(4)}\Bigg|_0^1 + \int_0^1 X_m{X_n}^{(4)}, && X_m(0) = X_n(1) = 0
\end{align}
$$ \implies \int_0^1 {X_m}^{(4)}X_n - \int_0^1 X_m{X_n}^{(4)} = 0 \implies (\beta_m^4 - \beta_n^4) \int_0^1 X_mX_n = 0 $$
since $\beta_m \ne \beta_n$, the integral must be $0$
