I have a eigen boundary value problem $$ \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2F=0 $$ $\mu$ is the separation variable or the eigenvalue variable with b.c. as $F(0)=0,\frac{F''(0)}{F'(0)}=\beta_h,\frac{F''(1)}{F'(1)}=\beta_h$
Typically the eigenvalue and eigen function relation is shown as
$$ op(\bar{f})=\mu \bar{f} $$ If the above relation is satisfied, then $\mu$ is called the eigenvalue of operator $op$ corresponding to eigen function $\bar{f}$
I am failing to express my EBVP in the standard form as my separation constant $\mu$ is attached to $F'$
ATTEMPT
$$ (\lambda_h\frac{d^2}{dx^2}-2\beta_h\lambda_h\frac{d}{dx}+{\beta_h}^2\int\mathbb{d}x)f=(-\lambda_h{\beta_h}^2+\beta_h+\mu)f $$
assuming $F:=\int f(x)$
but then i cannot form a characteristic equation to go about finding eigenvalues
NOTE
I must make a note here that the solution $F$ or $f$ would be used in the following function
$$ \theta_w=e^{-\beta_hx}F'(x)e^{-\beta_cy}G'(y)$$ or $$\theta_w=e^{-\beta_hx}f(x)e^{-\beta_cy}g(y)$$
The initial EBVP gets complicated because it is third order and my described attempt makes it a Integro - Differential equation. Stuck at this for a while now.
ATTEMPT 2
Substituted $F=e^{kx}$ into the original EBVP yields
$$ e^{kx}(\lambda_h k^3- 2 \lambda_h \beta_h k^2 + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) k + \beta_h^2)=0 $$ So the characteristic polynomial is $$\lambda_h k^3- 2 \lambda_h \beta_h k^2 + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) k + \beta_h^2=0$$
which can now have three roots