Solving the heat equation via Laplace Transform

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) \\ & u(x,0) = 0 \end{align}

Attempt:

So first, to make things a little nicer, I let $$u(y,t) = v(y,t) + \sin(t)$$. The system for $$v$$ is then given by

\begin{align} & \frac{\partial v}{\partial t} - \frac{\partial^2 v}{\partial y^2} = 0 \\ & v(0,t) = -\sin(t) \\ & v(\infty,t) = 0 \\ & v(x,0) = 0 \end{align}

Now I applied a Laplace Transform with respect to $$t$$. This gives

\begin{align} & p\bar v - \frac{d^2 \bar v}{dy^2} = 0 \\ & \bar v(0) = -\frac{1}{p^2+1} \\ & \bar v(\infty) = 0 \end{align}

Solving, this gives $$\bar v(y) = -\frac{1}{p^2+1}\exp(-\sqrt{p}y)$$.

So my question is, is what I have done so far correct?

And, assuming this is correct, how do you invert this to find $$v(y,t)$$?