= Solution
Put $D=\sigma^2>0$ and $\xi=x-x_0$. Logarithmic <differentiation> of the positive <Gaussian function> gives
$$
\phi_t=\left(-\frac1{2t}+\frac{\xi^2}{4Dt^2}\right)\phi,\qquad\phi_x=-\frac{\xi}{2Dt}\phi.
$$
Differentiating once more in space gives
$$
\phi_{xx}=\left(-\frac1{2Dt}+\frac{\xi^2}{4D^2t^2}\right)\phi.
$$
Multiplying by $D$ proves
$$
\boxed{\phi_t=D\phi_{xx}=\sigma^2\phi_{xx}\qquad(t>0).}
$$
The assumption $\sigma\ne0$ ensures both the normalization and exponential denominator are defined. This is the <heat kernel>, with <diffusion coefficient> $D$ and centre $x_0$.
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