= Solution
Insert the linear initial profile into the transformed <integral> from (b):
$$
u(x,t)=\frac a{\sqrt\pi}\int_{\mathbb R}
(x+2\sqrt t\,\eta)e^{-\eta^2}\,d\eta.
$$
The constant part integrates to $ax$. The other part is an absolutely integrable <odd function>, so its <integral> vanishes. Hence
$$
\boxed{u(x,t)=ax\qquad(t\geq0)}.
$$
As a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear <temperature> profile is stationary on the whole line. More generally, every <affine function> is preserved by this <heat-kernel solution>.
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