= Solution
{c}
Use the <Fourier transform> convention $\widehat u(s,t)=\int_{\mathbb R}e^{-isx}u(x,t)\,dx$. Then $\widehat{u_{xx}}=-s^2\widehat u$ and $\widehat{xu}=i\partial_s\widehat u$, so
$$
\widehat{\partial_x(xu)}=is(i\partial_s\widehat u)=-s\partial_s\widehat u.
$$
Thus the transformed <partial differential equation> and <initial condition> are
$$
\boxed{\partial_t\widehat u+s\partial_s\widehat u=-s^2\widehat u,\qquad\widehat u(s,0)=1.}
$$
The last equality is the <Fourier transform> of the <Dirac delta distribution>. The two terms in $\partial_x(xu)=u+xu_x$ must be transformed together; dropping $u$ would give the wrong equation.
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