Laplace transform on a Boundary value problem (ODE)with . Am i attempting correctly? From a system of PDEs where i used the following ansatz: $$\theta_w(x,y) = e^{-\beta_h x} f(x) e^{-\beta_c y} g(y)$$.  $F(x) := \int f(x) \, \mathrm{d}x$ and $G(y) := \int g(y) \, \mathrm{d}y$
So, $$\theta_w(x,y) = e^{-\beta_h x} F'(x) e^{-\beta_c y} G'(y)$$
I have the following two third order linear ODEs which have been arrived at after applying separation of variables
\begin{eqnarray}
 \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F &=& 0,\\
 V \lambda_c G''' - 2 V \lambda_c \beta_c G'' + \left( (\lambda_c \beta_c - 1) V \beta_c + \mu \right) G' + V \beta_c^2 G &=& 0,
\end{eqnarray}
$F$ is $F(x)$ and $G$ is $G(y)$. The boundary conditions are 
For $F$:
$$F(0)=0$$
$$\frac{F''(0)}{F'(0)}=\beta_h$$
$$\frac{F''(1)}{F'(1)}=\beta_h$$
$\lambda_h$, $\beta_h$ and $V$ are constants $>0$, while $\mu$ is the constant of separation. 
I decided to apply the Laplace transform to each ODE and find the functions $F$ and $G$ individually finally obtaining $\theta_w$.
Applying Laplace transform
($\bar F$ is $\mathcal{L}F(x)$)
$$\lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2=0$$
Taking constants before each term as $a$, $b$, $c$ and $d$ makes $a=\lambda_h$, $b=-2\lambda_h\beta_h$, $c=(\lambda_h\beta_h-1)\beta_h-\mu$ and $d={\beta_h}^2$, we have
$$a[s^3\bar F(s)-s^2 F(0)-sF'(0)-F''(0)]+b[s^2\bar F(s)-sF(0)-F'(0)]+c[s\bar F(s)-F(0)]+d[\bar F(s)]=0$$
I know that the seperation constant $\mu$ is contained in $c$ and is not a known quantity and can attain values $<,=,>0$. Ultimately after applying the first 2 b.c. i arrive at:
$$\bar F(s)[as^3+bs^2+cs+d]=F'(0)[s+b+a\beta_h]$$
I cannot proceed from here, as $F'(0)$ is not known. Also i have doubts whether my approach here to solve two variable separated ODEs with a separation constant using Laplace transform is correct in the first place or not ?
 A: I don't believe that the Laplace-transform approach will work well here, since not all conditions are given at the same point.
In general, the homogeneous third-order linear ODE with constant coefficients $a y''' + b y'' + c y' + d y = 0$ has four types of solutions, depending on the zeros of the characteristic polynomial $P(\lambda) = a \lambda^3 + b \lambda^2 + c \lambda + d$.
Such a cubic function has always one real zero, and the first distinction happens according to whether the other two zeros are real or complex.


*

*If all zeros are real, then the solution takes one of the forms
\begin{equation}
y = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} + C_3 e^{\lambda_3 x}, \quad y = (C_1 + C_2 x) e^{\lambda_1 x} + C_3 e^{\lambda_3 x}, \quad y = (C_1 + C_2 x + C_3 x^2) e^{\lambda_1 x},
\end{equation}
depending on the multiplicity of the zeros.

*If only one zero is real then the solution will be of the form
\begin{equation}
y = C_1 e^{\lambda_1 x} + e^{-b_1 x} \left( C_2 \cos(b_2 x) + C_3 \sin(b_2 x) \right).
\end{equation}
You could verify these solutions and then perhaps you could exclude some types using the given conditions. The configuration of the zeros of the characteristic polynomial $P$ will depend on the value of your separation constant, and you can distinguish these cases according to a discriminant similar as in the quadratic case (https://en.wikipedia.org/wiki/Cubic_function#The_discriminant), it's just more work!
Finally, this distinction should not be made according to the sign of the separation constant $\mu$ but rather according to the sign of the discriminant of the characteristic polynomial $P$, which depends on $\mu$.
