Solving partial differential equation using laplace transform with time and space variation I have a equation like this:

$\dfrac{\partial y}{\partial t} = -A\dfrac{\partial y}{\partial x}+ B \dfrac{\partial^2y}{\partial x^2}$

with the following I.C
 $y(x,0)=0$
and boundary conditions $y(0,t)=1$ and $y(\infty , t)=0$
I tried to solve the problem as follows:
Taking Laplace transform on both sides, 
$\mathcal{L}(\dfrac{\partial y}{\partial t}) = - A \mathcal{L}(\dfrac{\partial y}{\partial x})+B \mathcal{L}(\dfrac{\partial^2 y}{\partial x^2})$
Now, on the L.H.S we have, 
$sY-y(x,0)=sY$
$\mathcal{L}(\dfrac{\partial^2 y}{\partial x^2}) =\displaystyle \int  e^{-st} \dfrac{\partial^2 y}{\partial x^2} dt$
Exchanging the order of integration and differentiation 
$\displaystyle\mathcal{L}(\frac{\partial^2 y}{\partial x^2}) =\frac{\partial^2}{\partial x^2} \int  e^{-st} y(x,t) dt$
$\displaystyle\mathcal{L}(\frac{\partial^2 y}{\partial x^2}) =\frac{\partial^2}{\partial x^2}\mathcal{L}(y) = \frac{\partial^2Y}{\partial x^2} $
$\displaystyle\mathcal{L}\frac{\partial y}{\partial x} = \frac{\partial Y}{\partial x}$
Now, combing L.H.S and R.H.S, we have, 
$\displaystyle sY = - A \frac{\partial Y}{\partial x} + B \frac{\partial^2Y}{\partial x^2}$
Above equation might have three solutions:
If $b^2 - 4ac > 0 $ let $r_1=\frac{-b-\sqrt{b^2-4ac}}{2a}$ and $r_2 = \frac{-b+\sqrt{b^2-4ac}}{2a}$
The general solution is $\displaystyle y(x) = C_1e^{r_1x}+C_2 e^{r_2x}$
if  $b^2 - 4ac = 0 $, then the general solution is given by 
$ y(x)=C_1e^{-\frac{bx}{2a}}+C_2xe^-{\frac{bx}{2a}}$
if $b^2 - 4ac <0$ , then the general solution is given by
$y(x) = C_1e^{\frac{-bx}{2a}}\cos(wx) + C_2 e^{\frac{-bx}{2a}}\sin(wx)$
Since, A, and B are always positive in my problem, the first solution seems to be appropriate. 
Now, from this point I am stuck and couldn't properly use the boundary conditions. 
If anyone could offer any help that would be great. 
"Solution added"
The solution of the problem is 
$y(x,t)= \dfrac {y_0}{2} [exp(\dfrac {Ax}{B}erfc(\dfrac{x+At}{2\sqrt{Bt}}) + erfc(\dfrac{x-At}{2\sqrt{Bt}})$
 A: Using your notation, we have
$$B Y''(x,s) - A Y'(x,s) - s Y(x,s) = 0$$
$$Y(0,s)=\int_0^{\infty} dt\: 1 \cdot e^{-s t} = \frac{1}{s}$$
$$\lim_{x \rightarrow \infty} Y(x,s) = 0$$
The general solution to the equation is
$$Y(x,s) = M(s) e^{r_+ x} + N(s) e^{r_- x}$$
where
$$r_{\pm} = \frac{A \pm \sqrt{A^2+ 4 B s}}{2 B}$$
where
$$M(s) + N(s) = \frac{1}{s}$$
$$M(s)=0$$
The latter equation was determined by the limit at $\infty$.  (I am of course assuming that $B > 0$; the other case may be considered as well.)
Therefore
$$Y(x,s) = \frac{1}{s} \exp{\left[-\frac{\sqrt{A^2+4 B s}-A}{2 B} x \right ]}$$
It is this quantity that must be inverse LT'ed to get the solution to your equation, viz.
$$y(x,t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: \frac{1}{s} \exp{\left[-\frac{\sqrt{A^2+4 B s}-A}{2 B} x + s \, t\right ]}$$
A: I get a simpler procedure that without using laplace transform.
Note that this PDE is separable.
Let $y(x,t)=X(x)T(t)$ ,
Then $X(x)T'(t)=-AX'(x)T(t)+BX''(x)T(t)$
$X(x)T'(t)=(BX''(x)-AX'(x))T(t)$
$\dfrac{T'(t)}{T(t)}=\dfrac{BX''(x)-AX'(x)}{X(x)}=\dfrac{4B^2s^2-A^2}{4B}$
$\begin{cases}\dfrac{T'(t)}{T(t)}=\dfrac{4B^2s^2-A^2}{4B}\\BX''(x)-AX'(x)-\dfrac{4B^2s^2-A^2}{4B}X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^\frac{t(4B^2s^2-A^2)}{4B}\\F(x)=\begin{cases}c_1(s)e^\frac{Ax}{2B}\sinh xs+c_2(s)e^\frac{Ax}{2B}\cosh xs&\text{when}~s\neq0\\c_1xe^\frac{Ax}{2B}+c_2e^\frac{Ax}{2B}&\text{when}~s=0\end{cases}\end{cases}$
$\therefore y(x,t)=\int_{-\infty}^\infty C_1(s)e^\frac{2Ax+t(4B^2s^2-A^2)}{4B}\sinh xs~ds+\int_{-\infty}^\infty C_2(s)e^\frac{2Ax+t(4B^2s^2-A^2)}{4B}\cosh xs~ds$
$y(0,t)=1$ :
$\int_{-\infty}^\infty C_2(s)e^\frac{t(4B^2s^2-A^2)}{4B}~ds=1$
$C_2(s)=\dfrac{1}{2}\left(\delta\left(s-\dfrac{A}{2B}\right)+\delta\left(s+\dfrac{A}{2B}\right)\right)$
$\therefore y(x,t)=\int_{-\infty}^\infty C_1(s)e^\frac{2Ax+t(4B^2s^2-A^2)}{4B}\sinh xs~ds+\int_{-\infty}^\infty\dfrac{1}{2}\left(\delta\left(s-\dfrac{A}{2B}\right)+\delta\left(s+\dfrac{A}{2B}\right)\right)e^\frac{2Ax+t(4B^2s^2-A^2)}{4B}\cosh xs~ds=\int_{-\infty}^\infty C_1(s)e^\frac{2Ax+t(4B^2s^2-A^2)}{4B}\sinh xs~ds+e^\frac{Ax}{2B}\cosh\dfrac{Ax}{2B}$
