The problem under consideration is a linear convection PDE.

\(\) From Farlow’s “Partial Differential Equations for Scientists and Engineers”, chapter 15 problem 3 is $$u_t = -2u_x \quad -\infty < x < \infty \quad 0<t<\infty $$$$ u(x,0)=e^{-x^2}$$

Intuition

We see that the governing equation is a linear convection problem. The characteristic velocity is 2. We expect the solution to be a shifted initial condition $$u(x,t)=e^{-(x-2t)^{2}}$$

Laplace

The Laplace transform of the governing equation with respect to the temporal coordinate t  is $$L[u_{t}]=L[-2u_{x}]$$
Expanding with definitions gives $$U(x,s)-u(x,0)=-2\frac{dU(x,s)}{dx}$$
The transform of the IC is not found in Farlow’s tables $$L[u(x,0)]    =    L[e^{-x^{2}}]   =    ?$$
The Laplace ODE looks like the non-homogeneous equation $$\frac{dU(x,s)}{dx}+\frac{1}{2}sU(x,s)=\frac{1}{2}u(x,0)$$
Two difficult steps appear before we even see the inverse Laplace transform.
At this point, you should bail out and try another approach.
Fourier

The Fourier transform of the governing equation with respect to spatial coordinate x is $$F(u_{t})=F(-2u_{x})$$
Expanding with identities gives $$\frac{dU(t)}{dt}=-2i\zeta U(t)$$
The solution to U(t) , being a 1st order ODE, is $$U(t)    =    U(0)e^{-2i\zeta t}$$
Convolution $$u(x,t)    =    F^{-1}(U(0)e^{-2i\zeta t})
=    e^{-x^{2}}\star F^{-1}(e^{-2i\zeta t})$$
Finding $$e^{-2i\zeta t}$$
in Farlow’s table A gives $$f(x-a)$$
when $$a=2t$$

The solution is $$u(x,t)    =    e^{-(x-a)^{2}}
=    e^{-(x-2t)^{2}}$$