# Coordinate method to solve PDE

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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Take a look at this: eqworld.ipmnet.ru/en/solutions/fpde/fpdetoc1.htm – Shuhao Cao Sep 7 '12 at 18:10
Have a look at this. – Mhenni Benghorbal Sep 8 '12 at 2:36

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 $$\frac{1}{5}(-2 x^2-3 x y+2 y^2) - (x y - 2 x^2) = \frac{2}{5} (2 x - y)^2$$