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\varphi\psi'' = -c^2\varphi^{(4)}\psi\Rightarrow \frac{\psi''}{c^2\psi} = -\frac{\varphi^{(4)}}{\varphi} = -\beta^4.

$$

The spatial component is $\varphi^{(4)} - \beta^4\varphi = 0$. It is obvious that $\pm\beta$ are solutions to $m^4 - \beta^4 = 0$. That is, $(m - \beta)(m + \beta)(m^2 + \beta^2) = 0$. Thus, the finally two solutions are $m = \pm i\beta$.

$$

\varphi(x) = Ae^{\beta x} + Be^{-\beta x} + C\cos\beta x + D\sin\beta x

$$

which can be re-written as

$$

\varphi(x) = A\cosh\beta x + B\sinh\beta x + C\cos\beta x + D\sin\beta x.

$$

Using the boundary conditions, we have

$$

\varphi(0) = A + C = 0

$$

or $A = -C$.

$$

\varphi(x) = -C\cosh\beta x + B\sinh\beta x + C\cos\beta x + D\sin\beta x

$$

and

$$

\varphi'(0) = B\beta + D\beta = 0

$$

or $B = -D$.

So

$$

\varphi(x) = -C\cosh\beta x - D\sinh\beta x + C\cos\beta x + D\sin\beta x.

$$

For the remaining two boundary conditions, we have

$$

\varphi(L) = -C\cosh\beta L - D\sinh\beta L + C\cos\beta L + D\sin\beta L = 0.

$$

and

$$

\varphi''(L) = -C\beta^2\cosh\beta L - D\beta^2\sinh\beta L - C\beta^2\cos\beta L - D\beta^2\sin\beta L = 0.

$$

We can re-write the last boundary conditions in matrix form as

$$

\underbrace{\begin{pmatrix}

\cos\beta L - \cosh\beta L & \sin\beta L - \sinh\beta L\\

-\cos\beta L - \cosh\beta L & -\sin\beta L - \sinh\beta L

\end{pmatrix}}_{\text{matrix } A}

\begin{pmatrix}

C\\

D

\end{pmatrix} =

\begin{pmatrix}

0\\

0

\end{pmatrix}

$$

We will have a non trivial solution when the determinant of the matrix is 0.

\begin{alignat*}{3}

\det A & = & (\cos\beta L - \cosh\beta L)(-\sin\beta L - \sinh\beta L) - (-\cos\beta L - \cosh\beta L)(\sin\beta L - \sinh\beta L)\\

& = & -\cos\beta L\sin\beta L - \cos\beta L\sinh\beta L + \sin\beta L\cosh\beta L + \cosh\beta L\sinh\beta L\\

& & + \cos\beta L\sin\beta L - \cos\beta L\sinh\beta L + \sin\beta L\cosh\beta L - \cosh\beta L\sinh\beta L\\

& = & -2\cos\beta L\sinh\beta L + 2\sin\beta L\cosh\beta L\\

& = & \cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L

\end{alignat*}

That is, $\cos\beta L\sinh\beta L - \sin\beta L\cosh\beta L = 0$ for there to be a unique solution.

How do I the part below from the above?

Shown that the spatial eigenfunctions for these boundary conditions are given by

$$

X_n(x)\sim\sin\alpha_nx - \sinh\alpha_nx - \frac{\sin\alpha_nL - \sinh\alpha_nL}{\cos\alpha_nL - \cosh\alpha_nL}(\cos\alpha_nx - \cosh\alpha_nx)

$$