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 \end{align*}
 \ARROW To solve this equation we need to expand the $\nu(x,t)$ funciton in the Taylor series:
 \begin{align*}
-\nu(x,t) + \frac{\partial \nu(x,t)}{\partial t} \Delta t = p \nu(x,t) - p \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} p \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2\\ + q \nu(x,t) - q \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} q \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2
+\nu(x,t) + \frac{\partial \nu(x,t)}{\partial t} \Delta t = p \nu(x,t) - p \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} p \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2\\ + q \nu(x,t) + q \frac{\partial\nu(x,t) }{\partial x} \Delta x + \frac{1}{2} q \frac{\partial^2 \nu(x,t)}{\partial x^2} (\Delta x)^2
 \end{align*}
 \ARROW From which we get:
 \begin{align*}