Help factoring partial differential equation?

Question

Hi there, I have a question to answer which reads

"By factoring the differential operator, or otherwise, obtain the general solution of the following equation:

u_xx + u_xy - 2u_yy + 3u_x + 2u_y + 2u = 0", where u_xx means the second deriv of u with respect to x, or d^2u/dx^2 etc.

Can anyone explain how I go about factoring this? I've tried pretending the diff operator is a polynomial but I can't factor it at all. I'm assuming that since the question suggests factoring that it is actually possible.

Thanks,

There are no typos in the above PDE so it looks like it's unfactorable (unfortunately). Thanks

Answer 1

We want to factor it as
[(∂/∂x + a ∂/∂y + b) (∂/∂x + c ∂/∂y + d)] (u) = 0
which will gve us
[∂2/∂x^2 + c ∂2/∂x∂y + d ∂/∂x + a ∂2/∂x∂y + ac ∂2/∂y^2 + ad ∂/∂y + b ∂/∂x + bc ∂/∂y + bd] (u) = 0
i.e. u_xx + (a+c) u_xy + ac u_yy + (b+d) u_x + (ad + bc) u_y + bd u = 0
So we need
a+c = 1
ac = -2
b+d = 3
ad + bc = 2
db = 2

From the first two equations we get that a and c must be -1 and 2 in some order; from the third and fifth we get that b and d must be 1 and 2 in some order. Unfortunately, this means that ad + bc can only be 0 or 3, not 2. So unless you've made a typo this operator is not factorable.

Suppose the equation was
u_xx + u_xy - 2u_yy + 3u_x + 3u_y + 2u = 0
(3u_y instead of 2u_y). Then it would factor as
[(∂/∂x + 2 ∂/∂y + 1) (∂/∂x - ∂/∂y + 2)] (u) = 0.