= Solution
Work with $t>0$ and write $u=t^{-1/2}e^{-x^2/(4t)}$. The <chain rule> gives
$$
u_x=-\frac{x}{2t}u,\qquad u_{xx}=\left(\frac{x^2}{4t^2}-\frac1{2t}\right)u,\qquad u_t=\left(\frac{x^2}{4t^2}-\frac1{2t}\right)u.
$$
Thus $u_{xx}=u_t$: it is a constant multiple of the one-dimensional <Gaussian heat kernel>.
For the modified equation, put $f=g(t)u$. Substitution gives $g u_{xx}=g'u+gu_t-gu/t$. The heat-equation terms cancel, and $u>0$ permits division to obtain $\boxed{g'=g/t}$. Solving this <separable differential equation> gives
$$
\boxed{g(t)=Ct,\qquad f(x,t)=C\sqrt t\,e^{-x^2/(4t)},\quad t>0.}
$$
The zero solution is included by $C=0$. This is the <scalar integrating factor for a reaction-diffusion equation> with reaction coefficient $1/t$, because the modified equation is $f_t=f_{xx}+f/t$. Its sign produces growth. Indeed $\int_{\mathbb R}u\,dx=2\sqrt\pi$ is constant, while $\int_{\mathbb R}f\,dx=2C\sqrt\pi\,t$ increases linearly for $C>0$; neither profile is being assigned unit normalization here.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2-diffusion-and-growth.png]
{title=Diffusive spreading with conserved mass versus the same spreading with amplitude growth}
{description=Original profiles at $t=0.5,1,2$. The left solution is $t^{-1/2}e^{-x^2/(4t)}$; the right is $\sqrt t\,e^{-x^2/(4t)}$ with $C=1$. Their widths both grow as $\sqrt t$, but their spatial integrals are respectively constant and proportional to $t$.}
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