Semi-infinite heat/diffusion equation with B.C. and I.C. not equal to zero I need help for diffusion equation on semi-infinite rod. 
$$\frac{\partial u}{\partial t}=c\frac{\partial^2 u}{\partial x^2}$$
The initial condition and boundary conditions are:
$$u(x=0,t)=Y_s\\
u(x=\infty,t)=Y_0\\
u(x,t=0)=Y_0$$
And also boundary conditions are for $t\ge 0$; initial condition is for $~0\lt x\lt\infty$.
I tried to solve using separation of variables but couldn't, without having any boundary condition equal to zero. I would appreciate anyone to help me. Thank you in advance.
 A: Let $u=T(t)X(x)$, then $u_t=\frac{T'}{T}u$ and $u_{xx}=\frac{X''}{X}u$, so in the heat equation, for a constant $k$,
$$\frac{T'}{T}=k=c\frac{X''}{X}$$
Solving for T: We integrate $\frac{T'(t)}{T(t)}=k$ on both sides with respect to $t$, to get $\log(T(t))-log(T(0))=kt$ so $T(t)=Ae^{kt}$, for some constat $A$.
Solving for X: We have the equation $X''-\frac{k}{c}X=0$, which leads to the particular solution
$$X(x)=\begin{cases}
Bx+C&\text{if } \frac{k}{c}=0 \\
De^{\sqrt{\frac{k}{c}}x}+Ee^{-\sqrt{\frac{k}{c}}x}&\text{if } \frac{k}{c}>0 \\
F\sin(\sqrt{|\frac{k}{c}|}x)+G\cos(\sqrt{|\frac{k}{c}|}x)&\text{if } \frac{k}{c}<0
\end{cases}$$
Now, if $u(x=\infty, t)=Y_0<\infty$, this implies that, for $\frac{k}{c}=0$, we must have $B=0$, or else $u(x=\infty,t)=\pm \infty$ depending on $A>0$ or $A<0$. Also, if $\frac{k}{c}<0$, we must have $F=G=0$, as $\sin$ and $\cos$ are periodic functions without limit. If $\frac{k}{c}>0$: we need $D=0$ or else $u(\infty,0)\to\infty$. 
We have then two particular solutions: when $\frac{k}{c}=0$ and when $\frac{k}{c}>0$. The general solution must be a linear combination of them. When $\frac{k}{c}=0$, $X(x)=C$ and $T(t)=A$, so $u(x,t)=CA$. When $\frac{k}{c}>0$, $u(x,t)=Ee^{-\sqrt{\frac{k}{c}}x}$ and $T(t)=Ae^{kt}$, so $u(x,t)=Ae^{kt}Ee^{-\sqrt{\frac{k}{c}}x}$. Then, the general solution is 
$$u(x,t)=\alpha+\beta e^{-\sqrt{\frac{k}{c}}x+kt}$$
For some constants $\alpha$, $\beta$ and $k$, with $\frac{k}{c}>0$. With this solution, you can try to back out the constants, however, from the statement, It is not clear to me if $Y_0$ and $Y_s$ are constants or functions of $t$ and $x$. 
For instance, if $Y_0$ is a constant, $u(x=\infty,t)=Y_0$ implies that $\alpha=Y_0$, and so on.
