I'm trying to solve the equation
$$ \frac{\partial y}{\partial t} = \frac{\partial^2 y}{\partial x^2} $$
with conditions
$$y(x,0) = 0 \quad \textrm{for} \ 0 < x < 1$$
$$y(0,t) = 1$$ and $$\frac{\partial y}{\partial x}(1,t) = 0 \ .$$
I understand that the general approach for a case like this is to consider a steady state solution which has the inhomogenous boundary conditions, and then use separation of variables with homogenous boundary conditions to get the rest (which will, I believe, be a Fourier cosine series).
If the final boundary condition were $U(1,T) = 0$, then clearly $1 - X$ would be a suitable steady state solution.
However, I'm having a great deal of trouble coming up with anything suitable for this case that won't mess up the differential equation itself (i.e it needs to disappear when differentiated twice wrt $x$?)
I've tried thinking about things of the form $a x + b$, and also trigonometric possibilities, but they don't seem to work.
You were correct to start with
$$ y(x,t) = 1 + u(x,t) $$
We obtain the BVP for the remaining part:
\begin{cases} u_t = u_{xx} \\ u(0,t) = u_x(1,t) = 0 \\ u(x,0) = -1 \end{cases}
Separation of variables gives
$$ u(x,t) = \sum_{n=0}^\infty A_n \sin\left(\frac{2n+1}{2}\pi x\right)e^{-(2n+1)^2\pi^2t/4} $$
You'll need to prove the orthogonality of the $x$-functions before getting to the final solution.