4th order pde rotating bar $$u_{tt} − \omega^2_0  u + \lambda u_{xxxx} = 0,$$ 
where $\lambda$ is the wavelength (not the natural one). I've already separated the variables using $u(x,t)=f(x)g(t)$ and gotten here: 
$$\frac{g^{(2)}(t)}{g(t)}-w^2_0 + \lambda \frac{f^{(4)}(x)}{f(x)}=0.$$ 
There are initial and boundary conditions and stability questions but I'm really looking for the general solution to the pde since I can handle the rest.
 A: I don't think there is a neat general formula like for 1D wave equation. Separation of variables is the way to go, but boundary conditions are needed to determine the eigenfunctions $\phi_n$. From there, $$u(x,t) = \sum_n (A_n\cos \omega_n t+B_n\sin \omega_n t) \phi_n(x)$$ 
is the general form of $u$ with given boundary conditions. 
A: Let continue the separation of variables :
$$\frac{g^{(2)}(t)}{g(t)}-w^2_0 + \lambda \frac{f^{(4)}(x)}{f(x)}=0.$$
$$\begin{cases}
\frac{g^{(2)}(t)}{g(t)}=\alpha \\
\frac{f^{(4)}(x)}{f(x)}=\beta \\
\alpha-w^2_0+\lambda\beta=0
\end{cases}$$
$\alpha\:,\:\beta$ are any real or complex numbers.
Let $r_1(\alpha)$ and $r_2(\alpha)$ the two real or complex roots of $r^2=\alpha$
if $\alpha$ is negative or complex, sinusoidal terms will appears in the further solutions.
Let $R_1(\beta)\:,\:R_2(\beta)\:,\:R_3(\beta)\:,\:R_4(\beta)$ the four real or complex roots of $R^4=\beta$
$$\begin{cases}
g=c_1e^{r_1(\alpha)t}+c_2e^{r_2(\alpha)t} \\
f=C_1e^{R_1(\beta)x}+C_2e^{R_2(\beta)x} +C_3e^{R_3(\beta)x}+C_4e^{R_4(\beta)x} 
\end{cases}$$
The six coefficients are any real or complex values.
For each value of $\alpha$ and $\beta= \frac{w^2_0-\alpha}{\lambda}$ , a family of solutions of the ODE is :
$$u_\alpha(x,t)= \left( c_1(\alpha)e^{r_1(\alpha)t}+c_2(\alpha)e^{r_2(\alpha)t}  \right) \left(  C_1(\alpha)e^{R_1(\beta)x}+C_2(\alpha)e^{R_2(\beta)x} +C_3(\alpha)e^{R_3(\beta)x}+C_4(\alpha)e^{R_4(\beta)x} \right)$$
The general solution on the discret form is :
$$u(x,t)=\sum_{\text{any }\alpha} \left( \left( c_1(\alpha)e^{r_1(\alpha)t}+c_2(\alpha)e^{r_2(\alpha)t}  \right) \left(  C_1(\alpha)e^{R_1\frac{w^2_0-\alpha}{\lambda}x}+C_2(\alpha)e^{R_2\frac{w^2_0-\alpha}{\lambda}x} +C_3(\alpha)e^{R_3\frac{w^2_0-\alpha}{\lambda}x}+C_4(\alpha)e^{R_4\frac{w^2_0-\alpha}{\lambda}x} \right)  \right) $$
The general solution on the integral form is :
$$u(x,t)=\int \left( \left( c_1(\alpha)e^{r_1(\alpha)t}+c_2(\alpha)e^{r_2(\alpha)t}  \right) \left(  C_1(\alpha)e^{R_1\frac{w^2_0-\alpha}{\lambda}x}+C_2(\alpha)e^{R_2\frac{w^2_0-\alpha}{\lambda}x} +C_3(\alpha)e^{R_3\frac{w^2_0-\alpha}{\lambda}x}+C_4(\alpha)e^{R_4\frac{w^2_0-\alpha}{\lambda}x} \right)  \right) d\alpha$$
The coefficients are any derivable functions of $\alpha$.
The product of coefficients can be concatenated.
Note : This very complicated form of solution is due to the lack of boundary conditions. 
