TITLE: Canonical form of PDE
QUESTION [6 upvotes]: 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 $u(x,y) = x+f(x^2 - y) + g(y)$

REPLY [3 votes]: $$xu_{xx} + 2x^2u_{xy} -u_x =- 1$$
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
$$u_{\xi\eta}=0,$$
$$u_h=f(\xi)+g(\eta)=f(x^2-y)+g(y),$$
$$u=u_h+u_p=f(x^2-y)+g(y)+x$$