I am to find the nonpositive eigenvalues and eigenfunctions for the problem
$y^{(4)} + \lambda y = 0$
with the boundary conditions
$y(0) = y'(0) = 0$, $y(L) = y'(L) = 0$.
With some analysis, I can determine that $\lambda = 0$ is not an eigenvalue. However, for the case that $\lambda < 0$, I am having trouble determining the eigenvalues.
Using the general solution to the differential equation
$y_{gen} = c_1e^{\lambda^{\frac{1}{4}}x} + c_2e^{-\lambda^{\frac{1}{4}}x} + c_3\sin{(\lambda^{\frac{1}{4}}x)} + c_4\cos{(\lambda^{\frac{1}{4}}x)}$,
I should be able to determine a system of four equations for the constants. However, I become stuck at that point, and I am not sure the four equations are right
With $\lambda=-k^4 \land \lambda<0$ we get the general solution
$$y(x)=c_1 e^{k x}+c_2 e^{-k x}+c_3 \sin (k x)+c_4 \cos (k x)$$
For the given boundary conditions we get the equations $$y(0)=c_1+c_2+c_4=0$$ $$y'(0)=c_1 k-c_2 k+c_3 k=0$$ $$y(L)=c_1 e^{k L}+c_2 e^{-k L}+c_4 \cos (k L)+c_3 \sin (k L)=0$$ $$y'(L)=c_1 k e^{k L}-c_2 k e^{-k L}+c_3 k \cos (k L)-c_4 k \sin (k L)=0$$
We have a system of linear equations of the form $A\cdot c=0$ where $c=(c_1,c_2,c_3,c_4)^T$ with
$$A=\left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ k & -k & k & 0 \\ e^{k L} & e^{-k L} & \sin (k L) & \cos (k L) \\ k e^{k L} & k \left(-e^{-k L}\right) & k \cos (k L) & -k \sin (k L) \\ \end{array} \right)$$
$$det(A)=-4 k^2 (\cos (k L) \cosh (k L)-1)=0$$
The eigenvalues $k_{i}$ must satisfy $\cos (k_{i} L) \cosh (k_{i} L)-1=0$ in order to get a nontrivial solution.
Numerically we can find some eigenvalues depending on $L$, here $L=1$: $$k\approx (4.730040744862704,7.853204624095838,10.99560783800167,14.13716549125746,...)$$
We get the Matrix $B$ by substitution $B=A|_{\cos(k L)=\text{sech}(k L),\sin(k L)=\pm\sqrt{1-\text{sech}(k L)^2}}$
$$\implies det(B)=0, rank(B)=3$$
1st solution for $\sin(k L)=+\sqrt{1-\text{sech}(k L)^2}$:
$$\left\{c_2\to -\frac{c_1 e^{k L} \left(e^{k L}-\sqrt{1-\text{sech}^2(k L)}-\text{sech}(k L)\right)}{e^{k L} \sqrt{1-\text{sech}^2(k L)}-e^{k L} \text{sech}(k L)+1},c_3\to \frac{c_1 \left(e^{2 k L}-2 e^{k L} \text{sech}(k L)+1\right)}{-e^{k L} \sqrt{1-\text{sech}^2(k L)}+e^{k L} \text{sech}(k L)-1},c_4\to -\frac{c_1 \left(-e^{2 k L}+2 e^{k L} \sqrt{1-\text{sech}^2(k L)}+1\right)}{e^{k L} \sqrt{1-\text{sech}^2(k L)}-e^{k L} \text{sech}(k L)+1}\right\}$$
2nd solution for $\sin(k L)=-\sqrt{1-\text{sech}(k L)^2}$:
$$\left\{c_2\to \frac{c_1 e^{k L} \left(e^{k L}+\sqrt{1-\text{sech}^2(k L)}-\text{sech}(k L)\right)}{e^{k L} \sqrt{1-\text{sech}^2(k L)}+e^{k L} \text{sech}(k L)-1},c_3\to -\frac{c_1 \left(-e^{2 k L}+2 e^{k L} \text{sech}(k L)-1\right)}{e^{k L} \sqrt{1-\text{sech}^2(k L)}+e^{k L} \text{sech}(k L)-1},c_4\to -\frac{c_1 \left(e^{2 k L}+2 e^{k L} \sqrt{1-\text{sech}^2(k L)}-1\right)}{e^{k L} \sqrt{1-\text{sech}^2(k L)}+e^{k L} \text{sech}(k L)-1}\right\}$$
Now you may plug in the solution(s) into $y(x)$ and substitute $k=k_{i}$ and you have the eigenfunction of $k_{i}$ (1st solution applies to even $i$, 2nd solution applies to odd $i$).