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
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).