How to solve $U_{xy} + U_x + x + y = 1$ in a simple way?

Question

I have to solve the following PDE:

$$U_{xy} + U_x + x + y = 1$$

Additionally, I have the initial conditions $u(0,y) = 0$ and $u(x, 0) = 0$.

The reason why I am struggling with this in particular is, because it seems that there is a way to solve this without using the advanced procedures that are usually necessary to solve 2nd order PDEs. I tried to substitute with $v = u_x$ but it did not really get me to the solution (I have the result in Maple to check).

Any ideas?

Answer 1

Hint If we can find a particular solution $U^*(x, y)$, then writing the original p.d.e. in terms of $V = U - U^*$ yields the easier homogeneous p.d.e. $$V_{xy} + V_x = 0 .$$

Substituting a general quadratic ansatz, $$U^*(x, y) = A x^2 + B x y + C y^2 + D x + E y + F$$ gives $A = -\frac12, B = -1, D = 2,$ so taking $C = E = F = 0$ gives the particular solution $$U^*(x, y) = -\frac12 x^2 - x y + 2 x .$$

Additional Hint Writing the homogeneous p.d.e. in $W = V_x$ gives the o.d.e. $$W_y + W = 0,$$ so $$W = e^{-y} G(x)$$ for some function $G(x)$, hence the general solution to the homogeneous p.d.e. is $$V = e^{-y} G(x) + H(y).$$ Now it remains just to impose the boundary conditions on $$U = V + U^* = e^{-y} G(x) + H(y) -\frac12 x^2 - x y + 2 x .$$