$u''_{xy}+2xyu'_y-2xu=0.$
solve it for $u(x,y)$.
I received the following equations:
$u=\frac{1}{2x}v'_x+yv,$
$v''_{xy}+2xyv'_y=0.$
where $v=u'_y$. All my following tryings are worthless. I can't get the right answer, which is on this screenshot:
where $g$ and $f$ are arbitrary functions.
On
This is a partial answer.
$$u_{xy}+2xyu_y-2xu=0.$$ It is simple to check that $u=y$ is a particular solution. Using this one may seek solutions of the form $u=yf_1(x)+f_2(x)$. For $u$ of such a form one has $$u_y=f_1(x),\space\space u_{xy}=f_1'(x).$$ $f_1'$ stands for the derivative of $f_1$ wrt $x$. Upon substitution in the original equation $$f_1'(x)+2xyf_1(x)-2xyf_1(x)-2xf_2(x)=0.$$ It follows from above equation now that $$f_2(x)=\frac{f_1'(x)}{2x}$$ and one has solutions of the form $$u=yf(x)+\frac{f'(x)}{2x}.$$ The form of the original PDE also suggests that one may seek solutions of the form $$u=e^{-x^2}g(y).$$ For such a $u$, one has $$u_x=-2xe^{-x^2}g,\space\space u_y=e^{-x^2}g_y, \space\space u_{xy}=-2xe^{-x^2}g_y.$$ Substituting the above in the original equation, it is easy to see that $g=C(y-1)$ and one has solutions of the form $$u=Ce^{-x^2}(y-1).$$The last term in your solution is a more general form of above.
The formula (your screenshot) was probably obtained as shown below :
One typo corrected.