This is part 1 of problem 8 from the book "Computational Partial Differential equations Using MATLAB" by Jichun Li and Yi-Tung Chen:
Authors barely mention the method of separation of variables and that the solution is a linear combination of Fourier modes. So, after a lot of reading from different sources, I got this:
Let $u(x,t)$ be the solution of the given pde, this is the equation I'm looking for. Lets define the Fourier transform and its inverse for $u(x,t) as$:
$$U(w,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x,t)e^{-iwx}dx$$ $$u(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}U(w,t)e^{iwx}dw$$
Also I know that the fourier transform of $\frac{d^nu(x,t)}{(dx)^n}$ is $(iw)^nU(w,t)$.
Taking the fourier transform from both sides of the given pde: $$\frac{dU(w,t)}{(dt)} = (-w^2 - aiw)U(w,t)$$ $$U(w,t) = U(w,0) e^{-(w^2 + aiw)t}$$
Replacing $U(w,0)$ with the function from the fourier transform: $$U(w,t) = \frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\infty}u(x,0)e^{-iwx}dx) e^{-(w^2 + aiw)t}$$
Taking the inverse fourier transform: $$u(x,t)=\frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}u(x,0)dx) e^{-(w^2 + aiw)t}dw$$ $$u(x,t)=\frac{1}{2\pi}(\int_{-\infty}^{\infty}u(x,0)dx)\int_{-\infty}^{\infty} e^{-(w^2 + aiw)t}dw$$
Completing squares to get the normal distribution function (I'm operating on the exponentials): $$-(w^2 + aiw)t = -(w^2 + aiw + \frac{a^2}{2} - \frac{a^2}{2})t$$ $$= -(w^2 + aiw + \frac{a^2}{4})t - \frac{a^2t}{4}$$ $$= -(w - \frac{-ai}{2})^2t - \frac{a^2t}{4}$$
Puting the exponentials back to the original equation: $$u(x,t)=\frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\infty}u(x,0)dx)e^{- \frac{a^2t}{4}}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(w - \frac{-ai}{2})^2t}dw$$
$$u(x,t)=\frac{1}{\sqrt{4t\pi}}(\int_{-\infty}^{\infty}u(x,0)dx)e^{- \frac{a^2t}{4}}\frac{\sqrt{2t}}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(w - \frac{-ai}{2})^2t}dw$$
$$u(x,t)=\frac{e^{- \frac{a^2t}{4}}}{\sqrt{4t\pi}}(\int_{-\infty}^{\infty}u(x,0)dx)$$
$$\frac{u(x,t)}{dx}=\frac{e^{- \frac{a^2t}{4}}}{\sqrt{4t\pi}}u(x,0)$$
Let $k(t)$ be the boundary value when $x=0$ $$u(x,t)=k(t)\frac{e^{- \frac{a^2t}{4}}}{\sqrt{4t\pi}}e^{x}$$
My solution does not look like the one given by the book. I don't know how to proceed from this point.
Update: Following the comment of Joey Zou, I got:
From the value of $U(x,t)$: $$U(w,t) = \frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\infty}u(x,0)e^{-iwx}dx) e^{-(w^2 + aiw)t}$$
I can consider $u(x,t)$ some function $f(y)$ because it doesn't depend of $t$: $$U(w,t) = \frac{1}{\sqrt{2\pi}}(\int_{-\infty}^{\infty}f(y)e^{-iwy}dy) e^{-(w^2 + aiw)t}$$
Taking the inverse fourier transform: $$u(x,t)=\frac{1}{2\pi}(\int_{-\infty}^{\infty}f(y)dy)\int_{-\infty}^{\infty} e^{-w^2t - aiwt-iwy+iwx}dw$$
Update! Thanks Winther for noticing my mistakes:
Completing squares to get the normal distribution function: $$-w^2t - aiwt-iwy+iwx = -t[w^2 -(iwyt^{-1}-iwxt^{-1}-aiw) + (\frac{iyt^{-1}-ixt^{-1}-ai}{2})^2] + (\frac{iyt-ixt-ait}{2})^2$$
$$-w^2t - aiwt-iwy+iwx = -t[w^2 -(iwyt^{-1}-iwxt^{-1}-aiw) + (\frac{iyt^{-1}-ixt^{-1}-ai}{2})^2] - (\frac{yt-xt-at}{2})^2$$
Putting exponentials back to the equation: $$u(x,t)=\frac{1}{\sqrt{4t\pi}}(\int_{-\infty}^{\infty}f(y)dy)e^{- (\frac{yt-xt-at}{2})^2}\frac{\sqrt{2t}}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-(w - \frac{iyt^{-1}-ixt^{-1}-ai}{2})^2t}dw$$
$$u(x,t)=\frac{1}{\sqrt{4t\pi}}\int_{-\infty}^{\infty}f(y)e^{- (\frac{yt-xt-at}{2})^2}dy$$
All I know is that $u(x, 0) = f(y)$ but I do not have more information about the $y$ variable. So I'm stuck at this point.