Canonical form of PDE I want to solve the following PDE using canonical form:
$$xu_{xx} + 2x^2u_{xy} = u_x - 1$$
I have found the characteristic curves $\xi = y$ and $\eta = y-x^2$. By computing $u_{xx}$ and $u_{xy}$, I got the following (hopefully correct) relation :
$$4x^3u_{\xi\eta} = -1$$
How do I get the general solution from this?
The correct solution is $u(x,y) = x+f(x^2 - y) + g(y)$ 
 A: I have now solved it by putting $x^3 = (\xi - \eta)^{3/2}$ and then integrating the canonical equation will give the given solution.
Edit(Complete solution): 
Comparing the given PDE with the general form $A(x, y)u_{xx} + B(x, y)u_{xy} + C(x, y)u_{yy} = Φ(x, y,u,u_x,u_y)$ we find 
$$A = x, B= 2x^2, C = 0$$
Now, the characteristic curves can be found by the equation $\frac{dy}{dx} = \frac{B \pm\sqrt{B^2-4AC}}{2A} = 0, 2x$
Integrating, we find the 2 curves as $y = $constant and $y-x^2 = $constant, so we choose 
$$\xi = y,\qquad \eta = y-x^2$$ 
and apply the relations found by chain rule for differentiating $u$ w.r.t. $\xi$ and $\eta$:
$u_x = u_ξξ_x + u_ηη_x, u_y = u_ξξ_y + u_ηη_y,$
$u_{xx} = u_{ξξ}ξ^2_x + 2u_{ξη}ξ_xη_x + u_{ηη}η^2_x+ u_ξξ_{xx} + u_ηη_{xx}$
$u_{xy} = u_{ξξ}ξ_xξ_y + u_{ξη} (ξ_xη_y + ξ_yη_x) + u_{ηη}η_xη_y + u_ξξ_{xy} + u_ηη_{xy},$
$u_{yy} = u_{ξξ}ξ^2_y + 2u_{ξη}ξ_yη_y + u_{ηη}η^2_y + u_ξξ_{yy} + u_ηη_{yy}$
and substituting in original equation will give $4x^3u_{\xi\eta} = -1$
We want to eliminate $x$ so that equation consists only of $\xi$ and $\eta$, so put $$x^3 = (\xi - \eta)^{3/2}$$ then integrate equation once w.r.t. variable $\xi$ , then with $\eta$ to get the given solution.
A: $$xu_{xx} + 2x^2u_{xy} -u_x =- 1$$
Equation is linear. Then
$$u=u_h+u_p$$
where $u_p$ is is particular solution(we can take $\; u_p=x\;$),
$u_h$ is solution homogeneus equation $xu_{xx} + 2x^2u_{xy} -u_x =0$.
After change $\quad \xi = x^2-y,\quad \eta = y\quad$ we get
$$u_{\xi\eta}=0,$$
$$u_h=f(\xi)+g(\eta)=f(x^2-y)+g(y),$$
$$u=u_h+u_p=f(x^2-y)+g(y)+x$$
