Consider the following equation for $y(x)$ over $x\in[0, 1]$, where $\lambda\in \mathbb{C}$ is a parameter (the undetermined eigenvalue) $$xy''''+2y'''=\lambda y''$$ subject to $y(0)=y'(0)=y'(1)=0,~~y''(1)-y''(0)=\lambda\left(y(1)-y(0)\right) $.
Note: If one denote $f(x):=y''(x),$ then the equation becomes $$xf''+2f'=\lambda f.$$ It seems the solution of this equation belongs to some kind of hypergeometric functions, but I am not aware of what exactly the special function is.
(PS. My ultimate goal is to see whether there is an eigenvalue with $Re(\lambda)>0$.)
$$xf''+2f'+\lambda f=0$$ is a generalized form of Bessel differential equation : http://mathworld.wolfram.com/BesselDifferentialEquation.html
With the notations used in this page, Eqs.(3-4), $p=q=\gamma=\frac12$ and $\alpha=(-\lambda)^{1/2}$ . The solution in terms of Bessel functions is :
$$f(x)=C_1x^{-1/2} J_1\left(2(-\lambda x)^{1/2}\right) +C_2x^{-1/2} Y_1\left(2(-\lambda x)^{1/2}\right)$$ In case of $\lambda x<0$ , $J_1$ is the bessel function of first kind of order 1. $Y_1$ is the bessel function of second kind, order 1.
In case of $\lambda x>0$ , replace the Bessel functions $J_1\:,\:Y_1$ by the modified Bessel functions $I_1\:,\:K_1$ .
The integrations for $y(x)$ are certainely possible but arduous. They involve again Bessel functions : http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/21/01/02/01/01/02/