Trying to solve the PDE: $x^2\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}+xyu=1$ I tried to solve the following PDE:
$x^2\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}+xyu=1$, by a substitution of variables that would lead to an equation of the form
$\dfrac{\partial w}{\partial \xi}+h(\xi,\eta)w=F(\xi,\eta)$, but I fail to get to such equation.
I tried as the article suggested
$x^2\left(\dfrac{\partial w}{\partial \xi}\dfrac{\partial \xi}{\partial x} + \dfrac{\partial w}{\partial \eta}\dfrac{\partial \eta}{\partial x}\right) + y\left(\dfrac{\partial w}{\partial \xi}\dfrac{\partial \xi}{\partial y}+\dfrac{\partial w}{\partial \eta}\dfrac{\partial \eta}{\partial y}\right) + xyu=1$
and then got to $\dfrac{\partial w}{\partial \xi}\left( x^2 \dfrac{\partial \xi}{\partial x} + y\dfrac{\partial \xi}{\partial y}\right) + \dfrac{\partial w}{\partial \eta}\left( x^2 \dfrac{\partial \eta}{\partial x} + y \dfrac{\partial \eta}{\partial y}\right) + xyu = 1$
Imposing $ x^2 \dfrac{\partial \eta}{\partial x} + y \dfrac{\partial \eta}{\partial y} = 0$
I got to $\dfrac{\partial _x\eta}{\partial _y\eta}=-\dfrac{y}{x^2}=-\dfrac{dy}{dx}$
and
$\dfrac{1}{x} +ln y=\eta$
The article said that we could arbitraily choose $\xi$ as being $x$ but unfortunately I can't find it anymore.
Finally I've got to
$\xi^2\dfrac{\partial w}{\partial \xi}+\dfrac{\xi}{e^{\eta-\frac{1}{\xi^2}}}u=1$ but this $u$ seems out of place. As far as I remember, the equation shoud have become a solvable ODE at this point.
 A: $$x^2\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=1-xyu$$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dx}{x^2}=\frac{dy}{y}=\frac{du}{1-xyu}$$
A first characteristic equation comes from solving $\frac{dx}{x^2}=\frac{dy}{y}$ :
$$y\:e^{1/x}=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x^2}=\frac{du}{1-xyu}\quad\implies\quad \frac{dx}{x^2}=\frac{du}{1-x(c_1e^{-1/x})u}$
$\frac{du}{dx}+c_1\frac{1}{x}e^{-1/x}u=\frac{1}{x^2}\quad$ is a first order linear ODE which solution is :
$$u\:e^{-c_1\text{Ei}(-1/x)}-\int_1^x \left(e^{c_1\text{Ei}(-1/\xi)}\right)\frac{d\xi}{\xi^2}=c_2$$
Ei is the exponential integral.
The general solution of the PDE on the form of implicit equation $c_2=F(c_1)$ is :
$$u\:e^{-c_1\text{Ei}(-1/x)}-\int_1^x \left(e^{c_1\text{Ei}(-1/\xi)}\right)\frac{d\xi}{\xi^2}=F(c_1)$$
$$u=e^{c_1\text{Ei}(-1/x)}\left(\int_1^x \left(e^{c_1\text{Ei}(-1/\xi)}\right)\frac{d\xi}{\xi^2}+F(c_1)\right)$$
F is an arbitrary function, to be determined according to some boundary condition (missing in the wording of the question).
$$\boxed{u(x,y)=e^{y\:e^{1/x}\text{Ei}(-1/x)}\left(\int_1^x \left(e^{y\:e^{1/x}\text{Ei}(-1/\xi)}\right)\frac{d\xi}{\xi^2}+F(y\:e^{1/x})\right)}$$
