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?