Solve PDE $u_{xy}+xyu+yu_{x} + xu_{y}=0$ I need help to solve differential equation $u_{xy}+xyu+yu_{x} + xu_{y}=0$.
Equation and beginning of my solution is on photo.
Here $u_{y}$ is derivative with respect to $y$ etc.

 A: Another approach
$$
\left(\frac 1x\partial_x+1\right)\left(\frac 1y\partial_y+1\right)u = 0
$$
so now calling 
$$
U = \left(\frac 1y\partial_y+1\right)u
$$
we have
$$
\left(\frac 1x\partial_x+1\right)U = 0
$$
and solving for $U$ we have
$$
U(x,y) = e^{-\frac{x^2}{2}}\phi(y)
$$
etc.
A: Now by multiplying both sides in $e^{y^2\over 2}$ we obtain$$e^{y^2\over 2}u_y+yue^{y^2\over 2}=C(y)e^{y^2-x^2\over 2}\implies\\(e^{y^2\over 2}u)_y=C(y)e^{y^2-x^2\over 2}\implies \\e^{y^2\over 2}u=e^{-x^2\over 2}\int C(y)e^{y^2\over 2}dy+D(x)$$finally$$u=e^{-{x^2\over 2}}C(y)+e^{-{y^2\over 2}}D(x)$$which can alternatively be written as$$u=e^{-x^2\over2}e^{-y^2\over2}\left(e^{y^2\over 2}C(y)+e^{x^2\over2}D(x)\right)$$Now by defining $e^{x^2\over2}D(x)\to D(x)$ and  $e^{y^2\over 2}C(y)\to C(y)$ we have$$u=e^{-{x^2\over2}-{y^2\over2}}\left(D(x)+C(y)\right)$$
A: Another method (separation of variables). This is not the most elegant method. Just to not forget it.
One look for particular solutions on the form $\quad u=F(x)G(y)$.
$u_x=F'G\quad;\quad u_y=FG'\quad;\quad u_{xy}=F'G'$.
$$F'G'+xyFG+xf'G+yFG'=0$$
$$\left(\frac{F'}{xF}\right)\left(\frac{G'}{yG}\right)+1+\left(\frac{F'}{xF}\right)+\left(\frac{G'}{yG}\right)=0$$
This is possible only if $\left(\frac{F'}{xF}\right)=c_1$ and $\left(\frac{G'}{yG}\right)=c_2$ with $c_1c_2+1+c_1+c_2=0=(c_1+1)(c_2+1)=0.$
First particular solution with $c_1=-1$ :
$\left(\frac{F'}{xF}\right)=-1\quad\implies\quad X(x)=e^{-x^2/2}$  and $G(y)$ any function.
Second particular solution with $c_2=-1$ :
$\left(\frac{G'}{yG}\right)=-1\quad\implies\quad Y(y)=e^{-y^2/2}$  and $F(x)$ any function.
The solution of the ODE is any linear combination of the two particular solutions :
$$u(x,y)=e^{-x^2/2}G(y)+e^{-y^2/2}F(x)$$
