So there is a linear system
$ \frac{du}{dt} = - \left( \begin{array}{cc} 4 & -3 \\ 6 & -5 \end{array} \right) \frac{du}{dx} $
with an initial condition
$ u(x, 0)= \left( \begin{array}{c} -tanh(\lambda x) \\ 0 \end{array} \right) $
Now to start this off I have manipulated the first equation to get this
$ \left( \begin{array}{ccc} 1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\ 0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx} $
which can be rewritten as
$ \left( \begin{array}{ccc} 1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\ 0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx} =>\left( \begin{array}{cc} 1 & \frac{-3}{4}\\ 0 & \frac{1}{12}\end{array} \right) \frac{du}{dx} = \left( \begin{array}{c} -\frac{1}{4} \frac{du_{1}}{dt} \\ \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt} \end{array} \right) $
but I'm not sure how to deal with the derivatives. Would I use what I already have but just plug in the initial condition to $u_{1}$ and $u_{2}$ or am I using the wrong method?