<s>[INST] The eigenvalues and the eigenvectors of $y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0$ To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ I proceed as follows
$$\begin{align}
\bigg(\frac{d^{
4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{\lambda} \bigg) \bigg(\frac{d^{2}}{dt^{2}} - \sqrt{\lambda} \bigg) y\\
&= \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - i\lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} + \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - \lambda^{\frac{1}{4}} \bigg) y \\
&= 0
\end{align}$$
Therefore
$$\begin{align}y(t) &= A\cos(\lambda^{\frac{1}{4}}t)+B\sin(\lambda^{\frac{1}{4}}t)+ C\exp(\lambda^{\frac{1}{4}}t)+D\exp(-\lambda^{\frac{1}{4}}t)\\
y'(t) &= -\lambda^{\frac{1}{4}}A\sin(\lambda^{\frac{1}{4}}t)+ \lambda^{\frac{1}{4}}B\cos(\lambda^{\frac{1}{4}}t)+ \lambda^{\frac{1}{4}}C\exp(\lambda^{\frac{1}{4}}t)- \lambda^{\frac{1}{4}}D\exp(-\lambda^{\frac{1}{4}}t)
\end{align}$$
Let $v = \lambda^{\frac{1}{4}}$ then 
$$\begin{align}y(t) &= A\cos(vt)+B\sin(vt)+ C\exp(vt)+D\exp(-vt)\\
y'(t) &= -Av\sin(vt)+ Bv\cos(vt)+ Cv\exp(vt)- Dv\exp(-vt)
\end{align}$$
Now, applying the boundary conditions
$$\begin{align}y(0) &= 0 \implies A+C+D = 0 \\
y(1) &= 0 \implies A\cos(v)+B\sin(v)+ C\exp(v)+D\exp(-v)=0\\
y'(0) &= 0 \implies B+ C- D = 0\\
y'(1) &= 0 \implies -A\sin(v)+ B\cos(v)+ C\exp(v)- D\exp(-v)=0
\end{align}$$
Actually I stuck here.
 [/INST] This is my solution of the problem. Please let me know if you have any comments (Ahmed).
To find the eigenvalues and the eigenvectors of $$y^{(4)} = \lambda y,~y(0)=0,y(1)=0,y'(0)=0,y'(1)=0,$$ we proceed as follows
\begin{align*}
\bigg(D^4 - \lambda \bigg) y &= \bigg( D^2 + \sqrt{\lambda} \bigg) \bigg(D^2 - \sqrt{\lambda} \bigg) y\\
&= \bigg(D + i \sqrt[4]{\lambda} \bigg) \bigg(D - i\sqrt[4]{\lambda} \bigg) \bigg(D + \sqrt[4]{\lambda} \bigg) \bigg(D - \sqrt[4]{\lambda} \bigg) y \\
&= 0
\end{align*}
Therefore
\begin{align*}
y(x) &= A\cosh(\sqrt[4]{\lambda}x)+B\sinh(\sqrt[4]{\lambda}x)+ C\cos(\sqrt[4]{\lambda}x)+D\sin(\sqrt[4]{\lambda}x)\\
y'(x) &= \sqrt[4]{\lambda}A\sinh(\sqrt[4]{\lambda}x)+ \sqrt[4]{\lambda}B\cosh(\sqrt[4]{\lambda}x)- \sqrt[4]{\lambda}C\sin(\sqrt[4]{\lambda}x)+ \sqrt[4]{\lambda}D\cos(\sqrt[4]{\lambda}x)
\end{align*}
Let $r = \sqrt[4]{\lambda}$ then 
\begin{align*}
y(x) &= A\cosh(rx)+B\sinh(rx)+ C\cos(rx)+D\sin(rx)\\
y'(x) &= Ar\sinh(rx)+ Br\cosh(rx)- Cv\sin(rx)+ Dr\cos(rx)
\end{align*}
Now, applying the boundary conditions
\begin{align*}
y(0) &= 0 \implies A+C = 0 \\
y(1) &= 0 \implies A\cosh(r)+B\sinh(r)+ C\cos(r)+D\sin(r)=0\\
y'(0) &= 0 \implies B+D = 0\\
y'(1) &= 0 \implies r\left(A\sinh(r)+ B\cosh(r)- C\sin(r)+ D\cos(r)\right) =0
\end{align*}
So that $C = -A$ and $D = -B$. So that 
$$A(\cosh(r) - \cos(r)) + B(\sinh(r) - \sin(r))=0$$
$$A(\sinh(r) + \sin(r)) + B(\cosh(r) - \cos(r))=0$$
Therefore
$$
\begin{bmatrix}
\cosh(r) - \cos(r) & \sinh(r) - \sin(r) \\
\sinh(r) + \sin(r) & \cosh(r) - \cos(r)
\end{bmatrix} 
\begin{bmatrix}
A \\
B
\end{bmatrix} =
\begin{bmatrix}
0 \\
0
\end{bmatrix}
$$
From the basic theory of system of linear algebraic equations, compatibility requires the vanishing of the determinant of the matrix, i.e.,
$$\left(\cosh(r) - \cos(r)\right)^2 - \left(\sinh^2(r)-\sin^2(r) \right)=0$$
Hence
$$\cosh(r) \cos(r) = 1$$
Using numerical methods Maple finds the positive possible $r$'s to be $$r = [4.730, 7.853, 10.996, 14.137, \cdots ] \simeq \left[\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \cdots\right] = \frac{(2n+1)\pi}{2},~n=1,2, \cdots .$$ And since $r^4 = \lambda$, the possible approximate eigenvalues for the eigenvalue problem are 
$$\lambda = r^4 \simeq \frac{(2n+1)^4\pi^4}{16},~n=1,2, \cdots,$$
and the approximate eigenvectors are given by
\begin{align*}
y_n(x) \simeq &A_n \left(\cosh \frac{(2n+1)\pi}{2} x - \cos \frac{(2n+1)\pi}{2} x\right) \\
+& B_n \left(\sinh \frac{(2n+1)\pi}{2} x - \sin \frac{(2n+1)\pi}{2} x\right),~n=1,2, \cdots .
\end{align*}
</s>