Talk:Burgers' equation

Factor before the integral

It is $\ln(a\cdot b)=\ln(a)+\ln(b)$ and thus the factor $(4\pi \nu t)^{-1/2}$ could be splitted from the integral. The differentiation ${\frac {\partial }{\partial x}}$ then removes that summand. Therefore the solution is given by

$u(x,t)=-2\nu {\frac {\partial }{\partial x}}\ln {\Bigl \{}\int _{-\infty }^{\infty }\exp {\Bigl [}-{\frac {(x-x')^{2}}{4\nu t}}-{\frac {1}{2\nu }}\int _{0}^{x'}u(x'',0)dx''{\Bigr ]}dx'{\Bigr \}}$ --Hero Wanders (talk) 22:36, 14 March 2008 (UTC)