How to solve $u_{xy}+u_x+u_y=2$?

Question

How do I solve this equation $u_{xy}+u_x+u_y=2$ ? I know how to solve this $u_{xy}+u_y=0$ (with a change of variables and integrating factor $e^x$ ) but in this case another term was added $(u_x)$, should I follow the same idea or it requires a completely different method?

Answer 1

$$u_{xy}+u_x+u_y=2$$ Let $\quad u(x,y)=x+y+v(x,y)\quad$ which leads to : $$v_{xy}+v_x+v_y=0$$ Search for particular solutions of the form $v=X(x)Y(y)$

$X'Y'+X'Y+XY'=0\quad;\quad\frac{X'}{X}=-\frac{Y'}{Y+Y'}=\lambda \qquad\begin{cases} X=e^{\lambda x} \\ Y=e^{-\frac{\lambda}{1+\lambda}y}\end{cases}$

Particular solutions : $\quad v_\lambda(x,y)=e^{\lambda x-\frac{\lambda}{1+\lambda}y}$

The general solution is any linear combination of the particular solutions.

General solution expressed on the form of integral : $$v(x,y)=\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$ $f(\lambda)$ is an arbitrary function, in so far the integral be convergent. $$u(x,y)=x+y+\int f(\lambda)e^{\lambda x-\frac{\lambda}{1+\lambda}y}d\lambda$$ Or, equivalently : $$u(x,y)=x+y+\int g(\mu)e^{-\frac{\mu}{1+\mu}x+\mu y}d\mu$$ The function $f(\lambda)$ , or $g(\mu)$ , as well as the bounds of the integral, have to be determined according to the boundary conditions. This generally draw to solve an integral equation. Without well defined boundary conditions no further calculus is possible, even not to say if it is possible to analytically solve it and if there is a closed form for the solution.

Answer 2

Hint:

Let $u=v+x+y$ ,

Then $u_x=v_x+1$

$u_y=v_y+1$

$u_{xy}=v_{xy}$

$\therefore v_{xy}+v_x+1+v_y+1=2$

$v_{xy}+v_x+v_y=0$

Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,

Then $v_x=v_pp_x+v_qq_x=v_p+v_q$

$v_y=v_pp_y+v_qq_y=v_p-v_q$

$v_{xy}=(v_p+v_q)_y=(v_p+v_q)_pp_y+(v_p+v_q)_qq_y=v_{pp}+v_{pq}-v_{pq}-v_{qq}=v_{pp}-v_{qq}$

$\therefore v_{pp}-v_{qq}+v_p+v_q+v_p-v_q=0$

$v_{qq}=v_{pp}+2v_p$

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