Solving this PDE with boundary condition $u(\infty,t) = \sin(t)$ Question:
Let $u=u(y,t)$. Solve the following PDE (heat equation) in the region $y,t>0$:
\begin{align}
& \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial y^2} = \cos(t) \\
& u(0,t) = 0 \\
& u(\infty,t) = \sin (t)
\end{align}

Attempt:
I honestly have no idea where to start, since I don't really know how to deal with the boundary condition at infinity with the sine function.
I noticed that if I let $u(y,t) = v(y,t) + \sin(t)$, then we can get rid of the cosine in the PDE, since $v$ now satisfies
\begin{align}
& \frac{\partial v}{\partial t} - \frac{\partial^2 v}{\partial y^2} = 0 &&(1) \\
& v(0,t) = -\sin(t) &&(2) \\
& v(\infty,t) = 0 &&(3)
\end{align}
But I still don't know how to solve this. I know that
$$\frac{A}{\sqrt {t}}\exp\bigg(-\frac{(y-B)^2}{4t}\bigg)$$
satisfies $(1)$ and $(3)$, so am I supposed to superimpose these like a Fourier series?
I've also heard that you can solve this using a Fourier transform, but I don't know how that is possible since we only have a semi-infinite domain for both variables.
Any hints would be much appreciated. Thanks!
 A: $$ \frac{\partial u}{\partial t} - \frac{\partial^2 u}{\partial y^2} = \cos(t)$$
It is true that the change of function 
$$u(y,t)=v(y,t)+\sin(t)$$
transforms the PDE into :
$$ \frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial y^2} \qquad (1)$$
It is also true that the functions on the form
$$v(y,t)=\frac{A}{\sqrt {t}}\exp\bigg(-\frac{(y-B)^2}{4t}\bigg)$$
are solutions of Eq.$(1)$.
But why did you chose this particular form of solutions ? They are a lot of others which are likely to be linearly combined in order to satisfy the boundary conditions.
For example thanks to the method of separation of variables, one look for particular solutions on the form 
$$v(y,t)=T(t)Y(y)$$
$$T'(t)Y(y)=T(t)Y''(y)$$
$$\frac{T'}{T}=\frac{Y''}{Y}=\lambda$$
$$\begin{cases}
T(t)=c_1e^{\lambda t} \\
Y(y)=c_2e^{\pm\sqrt{\lambda}\:y }
\end{cases}\quad\implies\quad v=C\:e^{\lambda t\pm\sqrt{\lambda}\:y}$$
$\lambda$ is likely to be a complex number.
Any linear combination of these particular solutions is solution as well. So, a more general solution is 
$$v(y,t)=\sum_\lambda C_\lambda\:e^{\lambda t\pm\sqrt{\lambda}\:y}$$
We have to determine the values of $\lambda$ and $C_\lambda$ so that the boundary conditions be satisfied.
$$u(0,t)=0\quad\implies\quad v(0,t)=-\sin(t)\tag 2$$
$$u(\infty,t)=0\quad\implies\quad v(\infty,t)=0\tag 3$$
$$v(0,t)=\sum_\lambda C_\lambda\:e^{\lambda t}$$
Two terms are sufficient to satisfy $(2)$ with $\lambda=\pm i$ :
$v(0,t)=\frac{i}{2}e^{it}-\frac{i}{2}e^{-it}=\frac{i}{2}(\cos(t)+i\sin(t))-\frac{i}{2}(\cos(t)-i\sin(t)) = -\sin(t)$
$$v(y,t)=\frac{i}{2}e^{it\pm\sqrt{i}\:y}-\frac{i}{2}e^{-it\pm\sqrt{-i}\:y}$$
In order to satisfy $(3)$ the real part of $\sqrt{\pm i}$ must be negative. This determines the sign of the square root.
$$v(y,t)=\frac{i}{2}e^{it-\frac{1+i}{\sqrt{2}}y}-\frac{i}{2}e^{-it-\frac{1-i}{\sqrt{2}}y}$$
$v(y,t)=\frac{i}{2}e^{-\frac{1}{\sqrt{2}}y}\left(\cos(t-\frac{1}{\sqrt{2}}y)+i \sin(t-\frac{1}{\sqrt{2}}y)\right)-\frac{i}{2}e^{-\frac{1}{\sqrt{2}}y}\left(\cos(t-\frac{1}{\sqrt{2}}y)-i \sin(t-\frac{1}{\sqrt{2}}y)\right)$
$$v(y,t)=-e^{-\frac{1}{\sqrt{2}}y}\sin\left(t-\frac{1}{\sqrt{2}}y\right)$$
$$u(y,t)=\sin(t)-e^{-\frac{1}{\sqrt{2}}y}\sin\left(t-\frac{1}{\sqrt{2}}y\right)$$
