Using Laplace Transforms to solve a PDE I need to solve the following PDE, given the boundary conditions and initial condition, using Laplace Transforms:

$$ \begin{aligned} \displaystyle  K_1\frac{\partial^2u}{\partial x^2}+K_2u&=\frac{\partial u}{\partial t} \\ u(0,t)&=u_0 \\ u(\infty,t)&=0 \\ u(x,0)&=0 \end{aligned} $$

So far, I've taken the Laplace transform of the PDE to get the following result:
$$K_1\frac{\partial ^2U}{\partial x^2}+K_2U=sU$$
Solving the above ODE results in the following:
$$U(x,s)=C_1\exp\left(\sqrt{\frac{k_2+s}{k_1}}x\right)+C_2\exp\left(-\sqrt{\frac{k_2+s}{k_1}}x\right)$$
By taking the Laplace transform of the two boundary conditions, I get the following:
$$U(0,s)=\frac{u_0}{s}$$
$$U(\infty,s)=0$$
Using the second boundary condition, I can calculate that $C_2=0$, and that the PDE in terms of $x$ and $s$ is:
$$U(x,s)=\frac{u_0}{s}\exp\left(\sqrt{\frac{K_2+s}{K_1}}x\right)$$
Now, I have to take the inverse Laplace transform, but I'm having problems at this step.  I have the result which is:
$$ u(x,t)=\frac{2u_0}{\sqrt{\pi}}\int_{\frac{x}{2\sqrt{K_1t}}}^{\infty} \exp\left(-\lambda^2-\frac{K_2x^2}{4K_1\lambda^2}\right) \ d\lambda $$
It's suggested that I use one of the shifting theorems to obtain this solution, but I'm not sure how.  Thanks!
 A: There is some simplification if you rewrite the PDE as
$$
                   K_1\frac{\partial^2u}{\partial x^2}=\frac{\partial u}{\partial t}-K_2u \\
                     K_1 \frac{\partial^2 (e^{-K_2t}u)}{\partial^2 x}=\frac{\partial (e^{-K_2t}u)}{\partial t}
$$
Let $v(x,t)=e^{-K_2t}u(x,t)$. Then the new PDE becomes
$$
                v_t = K_1 v_{xx} \\
               v(0,t) = u_0e^{-K_2t} \\
               v(\infty,t) = 0 \\
               v(x,0) = 0.
$$
Now when you transform in $t$:
\begin{align}
              \mathscr{L}\{v_t\} & =\int_{0}^{\infty}v_t(x,t)e^{-st}dt \\
   & =v(x,t)e^{-st}|_{t=0}^{\infty}-\int_{0}^{\infty}v(x,t)e^{-st}(-s)dt \\
   & = s\mathscr{L}\{v\}.
\end{align}
So the transformed equations for $\mathscr{L}\{v\}(s,x)$ are
\begin{align}
    s\mathscr{L}\{v\}(s,x) &=K_1\mathscr{L}\{v\}_{xx}(s,x), \\
     \mathscr{L}\{v\}(s,0)&= \mathscr{L}\{u_0e^{-K_2t}\}=\frac{u_0}{s+K_2}.
\end{align}
The solution in $s,x$ is
$$
           \mathscr{L}\{v\}(s,x) = \frac{u_0}{s+K_2}\exp\left(-\sqrt{\frac{s}{K_1}}x\right).
$$
The inverse Laplace transform is a convolution in $t$ of inverse transforms of the two terms multiplying each other on the right above:
\begin{align}
    v(t,x) & = \left( u_0 e^{-K_2 t}\right)*\left( \frac{1}{\pi}\int_{0}^{\infty}e^{-st}\sin\left(\sqrt{\frac{s}{K_1}}x\right)ds\right) \\
   & = \frac{u_0}{\pi}\int_{0}^{\infty}\int_{0}^{t}e^{-K_2t'}e^{-s(t-t')}dt'\sin\left(\sqrt{\frac{s}{K_1}}x\right)ds
\end{align}
The integral in $t'$ is easily simplified and the solution $u(x,t)=e^{K_2t}v(x,t)$ of the original equation recovered, but I'll leave it there, before I compound my errors. :)
