Integration by parts results in $$ \frac{\partial \langle x^2\rangle}{\partial t} = -Dxw(x,t)|_{x=\pm\infty}+ 2D\int_{-\infty}^{\infty}w(x,t)dx=2D, $$ leading to $$ \langle x^2\rangle = 2Dt, $$ and the variance as $$ \begin{equation} \tag{6} \langle x^2\rangle-\langle x\rangle^2 = 2Dt. \end{equation} $$ The root mean square displacement after a time \( t \) is then $$ \begin{equation*} \sqrt{\langle x^2\rangle-\langle x\rangle^2} = \sqrt{2Dt}. \end{equation*} $$