# Math Help - numerical analysis

1. ## numerical analysis

2. 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.
$\frac{du}{dt}=Be^{Ax+Bt}$
$\frac{du}{dx}=Ae^{Ax+Bt}$
$\frac{d^2u}{dx^2}=A^2e^{Ax+Bt}$