I'm trying to solve $$\begin{cases} \dfrac{\partial u}{\partial t} = \dfrac{\partial^2 u}{\partial x^2}\ (x > 0,\ t > 0)\\ u(0,t)=1,\ u(x,0)=0 \end{cases}$$
I've transformed this to $$ sU - u(x,0) = \frac{d^2U}{dx^2} \to sU = \frac{d^2U}{dx^2} $$ But I don't know how could I anti-transform the resulting DE:
$$ U = A\exp(\sqrt{s}x) + B\exp(-\sqrt{s}x) $$
Thanks in advance.
A slightly more general problem is $$\begin{cases} \dfrac{\partial u}{\partial t} = k \, \dfrac{\partial^2 u}{\partial x^2} \hspace{5mm} (x > 0,\ t > 0)\\ u(0,t) = u_{0},\ u(x,0)=0. \end{cases}$$ Using the Laplace transform on the variable $t$ leads to $$ \bar{u}'' - \frac{s}{k} \, \bar{u} = 0 \hspace{5mm} \bar{u}(0, s) = \frac{u_{0}}{s}, \bar{u}(x,0)=0$$ and $$ \bar{u}(x, s) = A \, e^{\sqrt{s/k} \, x} + B \, e^{- \sqrt{s/k} \, x}.$$ The solution must be bounded as $x \to \infty$ which implies $A=0$. Applying the conditions yields $$ \bar{u}(x, s) = \frac{u_{0}}{s} \, e^{- \sqrt{s/k} \, x}. $$ Without proof of the inverse Laplace transform the solution is given by $$ u(x, t) = u_{0} \, \text{erfc}\left( \frac{x}{2 \, \sqrt{k \, t}}\right).$$