 Solution

Use the Fourier transform convention $\hat{u}(s,t)={\int }_{\mathbb{R}}{e}^{-isx}u(x,t)dx$. Then $\widehat{u_{xx}}=-s^2\hat{u}$ and $\widehat{xu}=i{\partial }_s\hat{u}$, soThus the transformed partial differential equation and initial condition areThe last equality is the Fourier transform of the Dirac delta distribution. The two terms in ${\partial }_x(xu)=u+x{u}_x$ must be transformed together; dropping $u$ would give the wrong equation.