Lagrange-Charpit equations: $$dx=\dfrac{dy}{2}=\dfrac{du}{(-2x+y)u+2x^2+3xy-2y^2}$$
$$\dfrac{dx}{dy}=\dfrac{1}{2} \implies x=\dfrac{y}{2}+A, A \in \mathbb{R}$$
By sagemath software a substitution for the variable $x$,
$$\dfrac{du}{dy}=\dfrac{(-2x+y)u+2x^2+3xy-2y^2}{2}=1.0 \, A^{2} - 1.0 \, A u + 2.5 \, A y$$
$$du=(A^2-Au+2.5Ay)dy$$
$$u=A^2y-A\int_L u(x,y)dy + \dfrac{5}{4}Ay^2 +B$$ Now I am stuck.
I solved by different method and obtained that $$u(x,y)=\frac{{\left(2 \, x - y\right)} {\left(x + 2 \, y\right)} - 5}{2 \, x - y} +C$$
How to proceed?
$$u_x + 2u_y + (2x − y)u = 2x^2 + 3xy − 2y^2$$ $$dx=\dfrac{dy}{2}=\dfrac{du}{(-2x+y)u+2x^2+3xy-2y^2}\qquad\text{is OK.}$$ First characteristic equation : $$x-\dfrac{y}{2}=A\qquad\text{is OK.}$$ Second characteristic equation : $$\dfrac{du}{dy}= A^{2} - A \: u(y) + \frac52A\:y \qquad\text{is OK.}$$ This is a first order linear ODE which solution is : $u=B\:e^{-Ay}+\frac52 y+A-\frac{5}{2A}$ $$\left(u-\frac52 y-A+\frac{5}{2A}\right)e^{Ay}=B$$ The general solution of the PDE on implicit form $B=F(A)$ is : $$\left(u-\frac52 y-(x-\dfrac{y}{2})+\frac{5}{2(x-\dfrac{y}{2})}\right)e^{(x-\dfrac{y}{2})y}=F\left(x-\dfrac{y}{2}\right)$$ $F$ is an arbitrary function (to be determined according to some boundary condition). Then, after simplification : $$u(x,y)=2y+x+\frac{5}{y-2x}+e^{(y-2x)y/2}F\left(x-\dfrac{y}{2}\right)$$