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I'm trying to solve the equation:

$ \frac{\partial U}{\partial x} + 2 \frac{\partial U}{\partial y} + (2x-y) U = 2x^2 + 3xy - 2y^2 $

I'm assuming this requires a change of variables but I'm not so sure about how to pick appropriately. An explanation of the procedure would be appreciated.

The answer is $ U = x + 2y + \frac{5}{y-2x} + e^{\frac{1}{5}(-2x^2 - 3xy + 2y^2)} f(y-2x) $

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    Have a look at [this](http://math.stackexchange.com/questions/$1$73532/coordinate-method-for-solving-first-order-linear-pde).2012-09-08

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If you write $V(x,t) = U(x,t+2x)$ (i.e. $y = t + 2x$), you get $\eqalign{\frac{\partial V}{\partial x} &= \dfrac{\partial U}{\partial x} + 2 \dfrac{\partial U}{\partial y}\cr &= - (2x-y) U(x,y) + 2 x^2 + 3 x y - 2 y^2\cr &= t V(x,t) - 5 x t - 2 t^2\cr}$ For fixed $t$ this is a first-order linear ODE in $x$, with solution $ V(x,t) = 2\,t+\frac{5}{t}+5\,x+{{\rm e}^{xt}}F(t)$ where $F$ is an arbitrary differentiable function. Substitute back $t=y-2x$ and you get $ U(x,y) = x + 2 y + \frac{5}{y-2x} + {\rm e}^{xy - 2 x^2} F(y-2x) $ Note that this is equivalent to the answer you wrote, because $ \frac{1}{5}(-2 x^2-3 x y+2 y^2) - (x y - 2 x^2) = \frac{2}{5} (2 x - y)^2 $