Reducing a first order linear pde to canonical form So the PDE is 
$\frac{\partial u}{\partial x} + 2 \frac{\partial u}{\partial y} + (2x-y)u = 2x^2 + 3xy -2y^2$ 
I worked out $c_1 = 2x- y$ and set this as $\xi = 2x - y$ and chose $\eta = x$ 
$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \xi} \frac{\partial \xi}{\partial x} + \frac{\partial u}{\partial \eta} \frac{\partial \eta}{\partial x}$ 
Therefore 
$\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial \xi} + \frac{\partial u}{\partial \eta}$
And 
$\frac{\partial u}{\partial y} = -\frac{\partial u}{\partial \xi}$
This transforms the original equation into 
$\frac{\partial u}{\partial \eta} + u \xi = 2x^2 + 3xy -2y^2$ 
Which is 
$\frac{\partial u}{\partial \eta} + u \xi = (2x-y)(x+2y)$ 
$\frac{\partial u}{\partial \eta} + u \xi = \xi (x+2y) $
I need to find the general solution for this PDE and am unsure how to continue on from here
 A: This is not an answer to the question about the canonical form because the hint was already given by Paul Sinclair in comment.
This is a comment about solving the PDE, but too long to be edited in comments section.
$$\frac{\partial u}{\partial x} + 2 \frac{\partial u}{\partial y} = -(2x-y)u + 2x^2 + 3xy -2y^2$$
The system of characteristic ODEs is :
$$\frac{dx}{1}=\frac{dy}{2}=\frac{du}{-(2x-y)u + 2x^2 + 3xy -2y^2}$$
A first characteristic equation comes from solving $\frac{dx}{1}=\frac{dy}{2}$ :
$$2x-y=c_1$$
A second characteristic equation comes from $\frac{dx}{1}=\frac{du}{-(2x-y)u + 2x^2 + 3xy -2y^2}$ with $y=2x-c_1$
$$\frac{du}{dx}=-\big(2x-(2x-c_1)\big)u + 2x^2 + 3x(2x-c_1) -2(2x-c_1)^2$$
$$\frac{du}{dx}=-c_1 u+5c_1 x-2(c_1)^2$$
This is a first order linear ODE easy to solve, leading to : 
$e^{c_1x}\left(-u+5x-\frac{5}{c_1}-2c_1 \right)=c_2$
$$e^{(2x-y)x}\left(-u+5x-\frac{5}{2x-y}-2(2x-y) \right)=c_2$$
$$e^{(2x-y)x}\left(-u+2y+x-\frac{5}{2x-y} \right)=c_2$$
The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is :
$$e^{(2x-y)x}\left(-u+2y+x-\frac{5}{2x-y} \right)=F(2x-y)$$
$F$ is an arbitrary function.
$$\boxed{u(x,y)=2y+x-\frac{5}{2x-y}-e^{(y-2x)x}F(2x-y)}$$ 
The function $F$ has to be determined in order to satisfy some boundary condition ( not specified in the wording of the question).
