Need explanation on how to use d’Alembert’s solution to solve and plot PDE's

Question:

 u(x,0) = e^{(-x^2)}  ,  u_t(x,0) = d/dx(e^{-x^2})

Solve  using d'Alembert's solution

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

$$u(x,0) = g(x) = e^{-x^2} \qquad
\boxed{
g(x-t) = e^{-(x-t)^2} \qquad
g(x+t) = e^{-(x+t)^2} }$$

$$u_t(x,0) = h(x) = \dfrac{d\left( e^{-x^2} \right)}{dx} \qquad
\boxed{
\int_{x-t}^{x+t} h(\xi) \, d\xi. = \int_{x-t}^{x+t} \dfrac{d\left( e^{-\xi^2} \right)}{d\xi} = \left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}\\
}$$

$$u(x,t) = \frac{1}{2}\left[g(x-t) + g(x+t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(\xi) \, d\xi.\\\\
u(x,t) = \frac{1}{2}\left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2}\left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}$$

$$u(x,t) = \frac{1}{2} \left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} - e^{-(x-t)^2} \right]\\\\
u(x,t) = \frac{1}{2} \left[ e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} } \right]\\\\
\boxed{ u(x,t) = e^{-(x+t)^2} }\\\\
u(x,0) = e^{-(x)^2}\\
u(x,1) = e^{-(x+1)^2}\\
u(x,10) = e^{-(x+10)^2}$$

41.024382993532i

Here f(x-at) = e^{-(x-t)^2} and g(s) = -2s*e^{-s^2} so:

I'll leave you to sketch the solutions.

(Note: for f use u(x,0), and for g use u_t(x,0)) 

Thanks a lot Alan, cant thumbs up enough =)

You're welcome!

I'll help with my thumb too :)

Thanks Alan and Heureka