Find the common solution of the PDE $u_{xy}+a(x,y)u_x=0$ 

Find the common solution of
    $$
u_{xy}+a(x,y)u_x=0~~~~~\text{in}~~~~~\Omega:=\left\{(x,y)\in\mathbb{R}^2 : \lvert x-x_0\rvert<d_1, \lvert y-y_0\rvert <d_2\right\},\\ 
d_1,d_2>0, (x_0,y_0)\in\mathbb{R}^2, a\in C(\Omega)
$$


In the meantime, I thought about an answer to this question (see below). Would be great to get a feedback to it.
With kind regards
 A: I add my answer in order to get a feedback from you. Please tell me, if my general solution is right or maybe nonsense, would be very kind of you!

First of all I use the theorem of Schwarz and substitute $z:=u_x$, getting an ODE of order 1 in $y$ (here $x$ is only a parameter):
$$
z_y=-a(x,y)z~~~~~(1)
$$
If one considers $a$ as only dependent on $y$ and $x$ only as a parameter, then $a$ is continious on the intervall $I:=(y_0-d_2,y_0+d_2)$. So the Fundamental Theorem of Calculus says that
$$
A(x):=\int_{y_0}^{y}a(x,\tau)\ d\tau~~~~~(2)
$$
is an antiderivative on $I$. So I continue (1) by separation of variables, getting
$$
z=C_1(x)\cdot\exp\left(-\int_{y_0}^{y}a(x,\tau)\ d\tau\right).
$$
Resubstituting, it follows
$$
u(x,y)=\int_{x_0}^{x}C_1(\eta)\exp\left(-\int_{y_0}^y a(\eta,\tau)\, d\tau\right)\, d\eta+C_2(y)~~~~~(3)
$$
with $C_1\in C^1((x_0-d_1,x_0+d_1))$ and $C_2\in C^1(I)$ arbitrary functions.
Comment: Instead of chosing $y_0$ and $x_0$ as lower borders in (2) and (3)it is of course possible to chose any point in $I$ respectively any point in $(x_0-d_1,x_0+d_1)$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\partiald{{\rm u}\pars{x,y}}{x}
&=
\exp\pars{-\int_{y_{0}}^{y}{\rm a}\pars{x,y'}\,\dd y'}
+
\partiald{{\rm u}\pars{x,y_{0}}}{x}
\\
\partiald{{\rm u}\pars{x,y_{0}}}{x}
&\equiv
\left.\partiald{{\rm u}\pars{x,y'}}{x}\right\vert_{y' = y_{0}}
\equiv
\tilde{\rm u}\pars{x,y_{0}}
\end{align}

\begin{equation}
{\rm u}\pars{x,y}
=
\int_{x_{0}}^{x}
\exp\pars{-\int_{y_{0}}^{y}{\rm a}\pars{x',y'}\,\dd y'}\,\dd x'
+
\int_{x_{0}}^{x}\tilde{\rm u}\pars{x',y_{0}}\,\dd x'
+
{\rm u}\pars{x_{0},y}
\tag{1}
\end{equation}

In addition,
$$
{\rm u}_{x}\pars{x,y}{\rm u}_{xy}\pars{x,y}
+
{\rm a}\pars{x,y}{\rm u}_{x}^{2}\pars{x,y}=0\,,
\quad\imp\quad
\bracks{{1 \over 2}\,\partiald{}{y} + {\rm a}\pars{x,y}}{\rm u}_{x}^{2}\pars{x,y} = 0
$$

$$
\partiald{}{y}\bracks{%
\exp\pars{2\int_{y_{0}}^{y}{\rm a}\pars{x,y'}\,\dd y'}
{\rm u}_{x}^{2}\pars{x,y}} = 0
$$
and
$$
{\rm u}_{x}^{2}\pars{x,y}
=
\tilde{\rm u}^{2}\pars{x,y_{0}}\exp\pars{-2\int_{y_{0}}^{y}{\rm a}\pars{x,y'}\,\dd y'}
$$

Let's assume we have two solutions $\phi\pars{x,y}$ and $\varphi\pars{x,y}$
of the differential equation for given values of ${\rm u}\pars{x_{0},y}$ and
$\tilde{\rm u}\pars{x,y_{0}}$. Then, we have $\delta_{x}^{2}\pars{x,y} = 0$
where $\delta\pars{x,y} \equiv \phi\pars{x,y} - \varphi\pars{x,y}$. We conclude
that
$$
\delta\pars{x,y} = {\rm f}\pars{y}\
\mbox{where}\ {\rm f}\ \mbox{is an }\ {\it arbitrary}\ \mbox{function}.
$$
However,
$$
{\rm f}\pars{y}
=
\phi\pars{x_{0},y} - \varphi\pars{x_{0},y}
=
{\rm u}\pars{x_{0},y} - {\rm u}\pars{x_{0},y} = 0
\quad\imp\quad
\phi\pars{x,y} = \varphi\pars{x,y}
$$
We conclude that the solution $\pars{1}$ is the unique solution for a given boundary conditions as defined by $\ds{{\rm u}\pars{x_{0},y}}$ and
$\ds{\tilde{\rm u}\pars{x_{0},y}
 \equiv
 \left.\partiald{{\rm u}\pars{x,y'}}{y'}\right\vert_{y'\ =\ y_{0}}}$.
