Find the canonical form of the elliptic equation $$u_{xx}-2u_{xy}+5u_{yy}=0.$$
Attempt. Since $a=1,~b=-2,~c=5$, we get $\displaystyle b^2-4ac=-16<0$, so the equation is indeed elliptic. The characteristic equation yields: $$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{b\pm i\sqrt{4ac-b^2}}{2a} = -1\pm 2i$$ and $y+1\pm 2i x=c$. Setting $ \xi(x,y):=y+1-2 ix,~~\eta(x,y):=y+1+2ix$ and $$\mu=\frac{\xi+\eta}{2} = y+1,~~~\nu=\frac{\xi-\eta}{2i} =-2x,$$
then the given equation becomes
$$Α(\mu,\nu)u_{\mu\mu}+B(\mu,\nu)u_{\mu\nu}+C(\mu,\nu)u_{\nu \nu} +D(\mu,\nu) u_{\mu}+E(\mu,\nu) u_{\nu}+F(\mu,\nu) u=G(\mu,\nu)$$
where
\begin{eqnarray} A(\mu,\nu) &=& a\mu_x^2+b\mu_x\mu_y+c\mu_y^2=5,\nonumber\\ B(\mu,\nu) &=& 2a\mu_x\nu_x+b(\mu_x\nu_y+\mu_y\nu_x)+2c\mu_y\nu_y=4,\nonumber\\ C(\mu,\nu) &=& a\nu_x^2+b\nu_x\nu_y+c\nu_y^2=4,\nonumber\\ D(\mu,\nu) &=& a\mu_{xx}+b\mu_{xy}+c\mu_{yy}+d\mu_x+e\mu_y=0,\nonumber\\ E(\mu,\nu) &=& a\nu_{xx}+b\nu_{xy}+c\nu_{yy}+d\nu_x+e\nu_y=0,\nonumber\\ F(\mu,\nu) &=& f =0 ~~and~~ G(\mu,\nu)~=~g=0.\nonumber \end{eqnarray}
Question: according to the general theory, the canonical form should be $u_{\mu\mu}+u_{\nu \nu}=H(\mu,\nu,u,u_\mu,u_\nu)$, which is not the case here. I cannot find the flaw, though.
Thanks for the help.
Making a change of variables such as
$$ \cases{\eta = x + \frac{\lambda-1}{5\lambda-1} y\\ \xi = x+\lambda y } $$
we obtain
$$ 4u_{\eta\eta}+(1-5\lambda)^2u_{\xi\xi}=0 $$
NOTE
The trick follows:
making the generic change of variables
$$ \cases{\eta = x + c_1 y\\ \xi = x+c_2 y } $$
we arrive at
$$ \left(5 c_2^2-2 c_2+1\right) u_{\xi\xi}+2 \left(c_1 \left(5 c_2-1\right)-c_2+1\right) u_{\eta\xi}+\left(5 c_1^2-2 c_1+1\right) u_{\eta\eta}=0 $$
then we choose $c_2 = \lambda, c_1 = \frac{\lambda-1}{5\lambda-1}$. Here $\lambda\in \mathbb{R},\lambda\ne \frac 15$.
Finally, choosing $(1-5\lambda)^2 = 4$ we have for $\lambda =\{ -\frac 15,\frac{3}{5}\}$
$$ 4u_{\eta\eta}+4u_{\xi\xi}=0\Rightarrow u_{\eta\eta}+u_{\xi\xi}=0 $$