# numerical analysis

• Nov 3rd 2009, 08:06 AM
nirvanakris
numerical analysis
• Nov 7th 2009, 02:20 PM
TriKri
Quote:

Originally Posted by nirvanakris
Well, since they have been so nice and given you the solution I guess you can assume that it will be on the form $u=e^{Ax+Bt}$, where A and B are constants.
$\frac{du}{dt}=Be^{Ax+Bt}$
$\frac{du}{dx}=Ae^{Ax+Bt}$
$\frac{d^2u}{dx^2}=A^2e^{Ax+Bt}$