Heat PDE with unusual, inhomogeneous Neumann boundary condition The ODE:
$$y_t=ky_{xx}$$
BCs:
$$y(0,t)=0\text{ and } y_x(L,t)=\alpha [y(L,t)+\beta]$$
So the latter is an inhomogeneous Neumann BC.
The domain:
$$0\leq x \leq L\text{ and }t\geq 0$$
An IC is also needed but not relevant to my question, right now.

I'm familiar with the method of *homogenisation* where a separate function is added to the target function so that the PDE and/or its BCs becomes homogeneous. That works very well in simple cases.
In accordance with that, for my first attempt, I assumed that:
$$y(x,t)=y_E(x)+z(x,t)$$
where $y_E(x)$ is the steady-state equation (so for $y_t=0$):
$$y_t=0\Rightarrow y_E''=0$$
$$\Rightarrow y_E(x)=c_1x+c_2$$
With $y(0,t)=0$:
$$\Rightarrow c_2=0$$
$$y_E'=c_1=\alpha [c_1L+\beta]$$
$$c_1=\alpha c_1+\alpha \beta$$
$$c_1=\frac{\alpha \beta}{1-\alpha L}$$
Recapping:
$$y_t(x,t)=z_t(x,t)$$
And:
$$y_{xx}(x,t)=z_{xx}(x,t)$$
And:
$$y_x(L,t)=\alpha [y(L,t)+\beta]$$
$$c_1 +z_x(L,t)=\alpha [c_1L+z(L,t)+\beta]$$
So homogenisation hasn't been achieved.
Any serious pointers would be much appreciated.
 A: The steady state solution for the original problem is $y_E(x)=\frac{\alpha \beta}{1-\alpha L} x$. The transient solution is given by $z(x,t)=y(x,t)-y_E(x)$ which now solves the PDE, for $(x,t)\in (0, L)\times (0,\infty)$,
$$z_t=k z_{xx},$$
and with BCs
$$z(0,t)=0 \,\text{  and  } \, z_x(L,t)-\alpha z(L,t)=0$$ for $t>0$ and IC $z(x,0)=g(x)-y_E(x)$ (where $g$ is the IC of the original problem, unspecified in the OP). Thus the $\beta$ term has vanished as indicated in my comment on the OP. The transient solution can then be found by separation of variables.
For then we get two ODEs one in $x$, $\phi''+\lambda^2 \phi = 0$ for $0<x<L$ with BC $\phi(0)=0$ and $\phi'(L)-\alpha \phi(L)=0$ and one in $t$, $T'+\lambda^2 k T=0$ with the IC. Solving the first ODE and imposing the first BC gives $\phi = c_2 \sin(\lambda x),$ and imposing the second BC and avoiding trivial solutions requires $\lambda$ to solve $\tan(\lambda L)=\lambda/\alpha$ which has infinite solutions $\lambda_n$ for $n\geq 1$. All together, we get
$$z(x,t)=\sum_{n} b_n \sin(\lambda_n x)e^{-\lambda_n^2 k t},$$
with initial condition
$$z(x,0)=\sum_n b_n \sin(\lambda_n x)=g(x)-y_E(x),$$
which leads to
$$b_n=\frac{\int_0^L [g(x)-y_E(x)]\sin(\lambda_n x) dx}{\int_0^L \sin^2(\lambda_n x) dx},$$
so that finally, returning back to $y=z+y_E$, we have
$$y(x,t)=\frac{\alpha \beta}{1-\alpha L} x + \sum_n b_n \sin(\lambda_n x) e^{-\lambda_n^2 k t},$$
where $\lambda_n$ and $b_n$ are defined above.

Please comment for clarifications or corrections.
A: The problem has a solution if we simplify the second BC slightly, so that $\beta=0$:
$$y(0,t)=0\text{ and } y_x(L,t)=\alpha y(L,t)$$
Carry out separation of variables, this will yield, with $-m^2$ the separation constant:
$$y_n(x,t)=A_n\exp(-\alpha m^2 t)\sin(mx)$$
Insert into:
$$y_x(L,t)=\alpha y(L,t)$$
$$-mA_n\cos(mL) =A_n\alpha\sin(mL)$$
$$\Rightarrow \tan(mL)=-\frac{m}{\alpha }$$
$$\mu=mL \Rightarrow$$
$$\tan \mu=-\frac{\mu}{\alpha L}$$
This is a transcendental equation which can be solved numerically for $\mu$.
But it is not a solution to the original problem.

Then I thought to carry out a substitution:
$$y(x,t)=u(x,t)-\beta$$
$$\Rightarrow u_x(L,t)=\alpha u(L,t)$$
But also:
$$u(0,t)=\beta$$
Snookered again!

Finally, changing the nature of the first BC to:
$$y_x(x,t)=0$$
Then with $y(x,t)=u(x,t)-\beta$:
$$y_x(x,t)=u_x(x,t)=0$$
This would yield:
$$\tan \mu=-\frac{\alpha L}{\mu}$$
But this too changes the nature of the original problem.
The key seems to be to eliminate $\beta$ while keeping the other BC homogeneous. But how?
