I'm supposed to reduce following polynomial to its canonical form. But my result differs from the one given in my book, so I'm not sure if it's correct too.
$$ q = u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0 $$
So, the characteristic quadratic polynomial is
$$ x^2 - xy -2y^2 $$
Here I'm using Lagrange's reduction method for quadratic polynomial:
\begin{align} x^2 - xy -2y^2 &= (x^2 - xy + \frac{1}{4}y^2) - \frac{1}{4}y^2 - 2y^2\\ &= (x-\frac{1}{2}y)^2 - \frac{9}{4}y^2\\ &= \xi^2 - \frac{9}{4}\eta^2\\\\ &\xi = x-\frac{1}{2}y\\ &\eta = y \end{align}
Now I can define the function $u$ like this:
$$ u(x,y) = u(\xi(x,y), \eta(x,y)) $$
Now I need to deduce $u_x$ and $u_y$ too:
\begin{align} u_x &= u_\xi \cdot \xi_x + u_\eta \cdot \eta_x = u_\xi\\ u_y &= u_\xi \cdot \xi_y + u_\eta \cdot \eta_y = -\frac{1}{2}u_\xi + u_\eta \end{align}
So, my canonical form looks like this: \begin{align} q &= u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + u_\xi - \frac{1}{2}u_\xi + u_\eta\\ &= \underline{\underline{u_{\xi\xi} - \frac{9}{4}u_{\eta\eta} + \frac{1}{2}u_\xi + u_\eta}} \end{align}
Is my solution correct or is there a mistake?
I have tried to reduce your characteristic quadratic polynomial and I got :
$$ Q(v)= -\frac18 (4y+x)^2+\frac 98 x^2$$ where v=(x,y) I hope it helps