Example 2. Determine the type of the equation

$$
u_{x x}-2 u_{x y}+u_{y y}+u_{x}-u_{y}+u=x y
$$

and reduce it to the canonical form.
Solution.

1. Determine the coefficients of the equation $a_{11}, a_{12}$, and $a_{22}$. We have

$$
a_{11}=1, \quad a_{12}=-1, \quad a_{22}=1
$$

2. Calculate the expression

$$
a_{12}^{2}-a_{11} a_{22}=1-1=0
$$

3. Since $a_{12}^{2}-a_{11} a_{22}=0$ for all $x, y$, equation (6) is of parabolic type in the entire $X O Y$ plane.
4. Find the general integrals of the characteristic equation:

$$
(d y)^{2}+2 d y d x+(d x)^{2}=0
$$

Solving this equation for $d y$, we get $d y=-d x$. Therefore, the characteristic equation (7) has a single integral

$$
h_{1}(x, y)=y+x
$$

5. In equation (6), make the change of variables:

$$
x, y \longmapsto \xi=h_{1}(x, y)=y+x, \quad \eta=y .
$$

(As $\eta$, we could have taken any other twice differentiable function that does not depend on $h_{1}(x, y)=y+x$.)

By the chain rule for differentiation:

$$
\begin{aligned}
& u_{x} \longmapsto u_{\xi}, \quad u_{y} \longmapsto u_{\xi}+u_{\eta}, \\
& u_{x x} \longmapsto u_{\xi \xi}, \quad u_{x y} \longmapsto u_{\xi \xi}+u_{\xi \eta}, \quad u_{y y} \longmapsto u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}
\end{aligned}
$$

As a result of the change of variables, equation (6) takes the form

$$
u_{\xi \xi}-2 u_{\xi \xi}-2 u_{\xi \eta}+u_{\xi \xi}+2 u_{\xi \eta}+u_{\eta \eta}+u_{\xi}-u_{\xi}-u_{\eta}+u=(\xi-\eta) \eta
$$

i.e., equation (6) is reduced to the canonical form

$$
u_{\eta \eta}=u_{\eta}-u+(\xi-\eta) \eta
$$

Answer. Equation (6) is of parabolic type in the entire $X O Y$ plane and has the canonical form

$$
u_{\eta \eta}=u_{\eta}-u+(\xi-\eta) \eta
$$

where $\xi=y+x$ and $\eta=y$.