I need to use the Laplace Transform to solve the following PDE, but I don't think I'm doing it correctly. $u_{t}(y,t)=\nu\nabla^2 u(y,t)$ with $u(0,t)=u_{0}$ and $u(y,0)=0$.
What I have so far: Taking the Laplace Transform of both sides gives $s\tilde{u}(y,s)=\nu\tilde{u_{yy}}$ where $\tilde{u}$ denotes the Laplace transform of $u$. This ODE has solution $\tilde{u}(y,s)= Aexp(\sqrt{\frac{s}{\nu}y}) + Bexp(-\sqrt{\frac{s}{\nu}y})$. However I'm unsure where to go from here, this doesn't seem right to me so any help will be appreciated!
Coming back to this, I wanted to say that this problem can actually be solved by doing an odd reflection through the origin, as you appear to be trying to solve the heat equation in the half open domain $[0,\infty)$. Consider first letting $v(y,t) = u(y,t) - u_0$. Here I'll be assuming $u_0 \in \mathbb{R}$ since you didn't specify. Then our new $v$ can solve
\begin{cases} v_{t} - \nu \Delta v=0 & \text{in } (0,\infty) \times (0,\infty) \\ v(y,0) = -u_{0} & \text{on } (0,\infty) \times \{ t=0 \}\\ v(0,t) = 0 & \text{on } \{ y = 0 \} \times (0,\infty) \end{cases}
Now, define
$ \phi(y) = \begin{cases} -u_0 & y>0 \\ u_0 & y<0 \\ 0 & y =0 \end{cases} $
Then we can solve the corresponding IVP
\begin{cases} v_{t} - \nu \Delta v=0 & \text{in } \mathbb{R} \times (0,\infty) \\ v(y,0) = \phi(y) & \text{on } \mathbb{R} \times \{ t=0 \}\\ \end{cases}
Which you should be able to solve via a Laplace Transform pretty easily.