= Solution
{c}
Substitute $\widehat u=e^{-b(t)s^2}$ into the transformed <partial differential equation>. Comparing coefficients gives $b'=1-2b$, $b(0)=0$, whence
$$
\boxed{b(t)=\frac{1-e^{-2t}}2,\qquad\widehat u(s,t)=\exp\!\left[-\frac{1-e^{-2t}}2s^2\right].}
$$
The stated Gaussian <Fourier transform> pair, with its inverse factor $1/(2\pi)$, gives
$$
\boxed{u(x,t)=\frac{1}{\sqrt{2\pi(1-e^{-2t})}}\exp\!\left[-\frac{x^2}{2(1-e^{-2t})}\right],\qquad t>0.}
$$
This is a <normal distribution> density with mean zero and <variance> $1-e^{-2t}$. Its total integral is one. As $t\downarrow0$ it converges to $\delta$ in the sense of <distributions>, while as $t\to\infty$ it tends to the standard <normal distribution> density. This is the transition density from zero of the <Ornstein-Uhlenbeck process> with drift $-x$ and diffusion coefficient $\sqrt2$.
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