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.
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*}