Find the eigenvalues and eigenfunctions of the problem $$y^{(4)} − λy = 0$$
with the boundary conditions
(i) $\quad y(0) = y'' (0) = y(β) = y'' (β) = 0$
(ii) $\quad y(0) = y' (0) = y'' (β) = y''' (β) = 0$
(iii) $\quad y(0) = y' (0) = y(β) = y' (β) = 0$
$$\begin{align} \bigg(\frac{d^{4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{\lambda} \bigg) \bigg(\frac{d^{2}}{dt^{2}} - \sqrt{\lambda} \bigg) y\\ &= \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - i\lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} + \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - \lambda^{\frac{1}{4}} \bigg) y \\ &= 0 \end{align}$$
One of the solutions is given by
$$\begin{align} \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) y &= \frac{dy}{dt} + i \lambda^{\frac{1}{4}} y \\ &= 0 \\ \implies y_1(t) &= A\exp(-i\lambda^{\frac{1}{4}}t) \end{align}$$
where $A$ is a constant. Now it remains to solve each of the other three ODEs (the bracketed terms above), $y_2, y_3, y_4$ to get your solution. Remember to add the solutions together for the total solution $y(t) = y_1 + y_2 + y_3 + y_4$, then use your BCs to solve for the constants $A, B, C, D$.