How to solve second order PDE with first order terms. I know we can transform a second order PDE into three standard forms. But how to deal with the  remaining first order terms?
Particularly, how to solve the following PDE:
$$
u_{xy}+au_x+bu_y+cu+dx+ey+f=0
$$
update:
$a,b,c,d,e,f$ are all constant.
 A: Case $1$: $a=b=c=0$
Then $u_{xy}+dx+ey+f=0$
$u_{xy}=-dx-ey-f$
$u_x=\int(-dx-ey-f)~dy$
$u_x=C(x)-dxy-\dfrac{ey^2}{2}-fy$
$u=\int\left(C(x)-dxy-\dfrac{ey^2}{2}-fy\right)dx$
$u=C_1(x)+C_2(y)-\dfrac{dx^2y}{2}-\dfrac{exy^2}{2}-fxy$
Case $2$: $a\neq0$ and $b=c=0$
Then $u_{xy}+au_x+dx+ey+f=0$
Let $u_x=v$ ,
Then $u_{xy}=v_y$
$\therefore v_y+av+dx+ey+f=0$
$v_y+av=-dx-ey-f$
$(\exp(ay)v)_y=-dx\exp(ay)-ey\exp(ay)-f\exp(ay)$
$\exp(ay)v=\int(-dx\exp(ay)-ey\exp(ay)-f\exp(ay))~dy$
$\exp(ay)u_x=C(x)-\dfrac{dx\exp(ay)}{a}-\dfrac{ey\exp(ay)}{a}+\dfrac{e\exp(ay)}{a^2}-\dfrac{f\exp(ay)}{a}$
$u_x=C(x)\exp(-ay)-\dfrac{dx}{a}-\dfrac{ey}{a}+\dfrac{e}{a^2}-\dfrac{f}{a}$
$u=\int\left(C(x)\exp(-ay)-\dfrac{dx}{a}-\dfrac{ey}{a}+\dfrac{e}{a^2}-\dfrac{f}{a}\right)dx$
$u=C_1(x)\exp(-ay)+C_2(y)-\dfrac{dx^2}{2a}-\dfrac{exy}{a}+\dfrac{ex}{a^2}-\dfrac{fx}{a}$
Case $3$: $b\neq0$ and $a=c=0$
Then $u_{xy}+bu_y+dx+ey+f=0$
Let $u_y=v$ ,
Then $u_{xy}=v_x$
$\therefore v_x+bv+dx+ey+f=0$
$v_x+bv=-dx-ey-f$
$(\exp(bx)v)_x=-dx\exp(bx)-ey\exp(bx)-f\exp(bx)$
$\exp(bx)v=\int(-dx\exp(bx)-ey\exp(bx)-f\exp(bx))~dx$
$\exp(bx)u_y=C(y)-\dfrac{dx\exp(bx)}{b}+\dfrac{d\exp(bx)}{b^2}-\dfrac{ey\exp(bx)}{b}-\dfrac{f\exp(bx)}{b}$
$u_y=C(y)\exp(-bx)-\dfrac{dx}{b}+\dfrac{d}{b^2}-\dfrac{ey}{b}-\dfrac{f}{b}$
$u=\int\left(C(y)\exp(-bx)-\dfrac{dx}{b}+\dfrac{d}{b^2}-\dfrac{ey}{b}-\dfrac{f}{b}\right)dy$
$u=C_1(x)+C_2(y)\exp(-bx)-\dfrac{dxy}{b}+\dfrac{dy}{b^2}-\dfrac{ey^2}{2b}-\dfrac{fy}{b}$
Case $4$: $a,b,c\neq0$
Then $u_{xy}+au_x+bu_y+cu+dx+ey+f=0$
Try let $u=p(x)q(y)v$ ,
Then $u_x=p(x)q(y)v_x+p_x(x)q(y)v$
$u_y=p(x)q(y)v_y+p(x)q_y(y)v$
$u_{xy}=p(x)q(y)v_{xy}+p(x)q_y(y)v_x+p_x(x)q(y)v_y+p_x(x)q_y(y)v$
$\therefore p(x)q(y)v_{xy}+p(x)q_y(y)v_x+p_x(x)q(y)v_y+p_x(x)q_y(y)v+a(p(x)q(y)v_x+p_x(x)q(y)v)+b(p(x)q(y)v_y+p(x)q_y(y)v)+cp(x)q(y)v+dx+ey+f=0$
$p(x)q(y)v_{xy}+p(x)(q_y(y)+aq(y))v_x+(p_x(x)+bp(x))q(y)v_y+(p_x(x)q_y(y)+ap_x(x)q(y)+bp(x)q_y(y)+cp(x)q(y))v=-dx-ey-f$
Take $q_y(y)+aq(y)=0\Rightarrow q(y)=\exp(-ay)$ and $p_x(x)+bp(x)=0\Rightarrow p(x)=\exp(-bx)$ , the PDE becomes
$\exp(-bx-ay)v_{xy}+(c-ab)\exp(-bx-ay)v=-dx-ey-f$
$v_{xy}+(c-ab)v=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$
Case $4a$: $c=ab$
Then $v_{xy}=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$
$v_x=\int(-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay))~dy$
$v_x=C(x)-\dfrac{dx\exp(bx+ay)}{a}-\dfrac{ey\exp(bx+ay)}{a}+\dfrac{e\exp(bx+ay)}{a^2}-\dfrac{f\exp(bx+ay)}{a}$
$v=\int\left(C(x)-\dfrac{dx\exp(bx+ay)}{a}-\dfrac{ey\exp(bx+ay)}{a}+\dfrac{e\exp(bx+ay)}{a^2}-\dfrac{f\exp(bx+ay)}{a}\right)dx$
$\exp(bx+ay)u=C_1(x)+C_2(y)-\dfrac{dx\exp(bx+ay)}{ab}+\dfrac{d\exp(bx+ay)}{ab^2}-\dfrac{ey\exp(bx+ay)}{ab}+\dfrac{e\exp(bx+ay)}{a^2b}-\dfrac{f\exp(bx+ay)}{ab}$
$\exp(bx+ay)u=C_1(x)+C_2(y)-\dfrac{(dx+ey+f)\exp(bx+ay)}{ab}+\dfrac{d\exp(bx+ay)}{ab^2}+\dfrac{e\exp(bx+ay)}{a^2b}$
$u=C_1(x)\exp(-bx-ay)+C_2(y)\exp(-bx-ay)-\dfrac{dx+ey+f}{ab}+\dfrac{d}{ab^2}+\dfrac{e}{a^2b}$
$u=C_1(x)\exp(-ay)+C_2(y)\exp(-bx)-\dfrac{dx+ey+f}{ab}+\dfrac{d}{ab^2}+\dfrac{e}{a^2b}$
Hence the really difficult case is that when $c\neq ab$ . By letting $u=\exp(-bx-ay)v$ the PDE will reduce to $v_{xy}+(c-ab)v=-dx\exp(bx+ay)-ey\exp(bx+ay)-f\exp(bx+ay)$ , which is as headache as https://math.stackexchange.com/questions/218425 for finding its most general solution.
