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$u''_{xy}+2xyu'_y-2xu=0.$

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:

enter image description here

where $g$ and $f$ are arbitrary functions.

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  • $\begingroup$ what's the meaning of the subindeces $_x$ and $_y$? Are they derivatives? $\endgroup$ – seoanes Mar 20 '16 at 14:49
  • $\begingroup$ @seoanes Yes, of course. I'm sorry, it's a habit... $\endgroup$ – Lust_For_Love Mar 20 '16 at 14:52
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The formula (your screenshot) was probably obtained as shown below :

enter image description here

One typo corrected.

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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.

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