I Partial Differential Equation solved using Products

[mathhabibi]

TL;DR Summary

The following PDE$$P^2\frac{\partial^2P}{\partial x\partial y}=(P^2+1)\frac{\partial P}{\partial x}\frac{\partial P}{\partial y}+P\frac{\partial^2P}{\partial y\partial x}$$can be solved with products, my question is how to solve without knowing their properties.

Using the concepts of Summability Calculus but generalized such that the lower bound for sums and products is also variable, we can prove that the solution to the following PDE: $$P^2\frac{\partial^2P}{\partial x\partial y}=(P^2+1)\frac{\partial P}{\partial x}\frac{\partial P}{\partial y}+P\frac{\partial^2P}{\partial y\partial x}$$subject to the initial curve $P(a-1,a)=1$ where $P=P(x,y)$ is $P=\prod_{k=y}^xf(k)$, where $f(x)$ is assumed to be a function defined on the integers. The author goes over different methods of calculating sums of non-integer upper and lower bounds, which can be modified to products as well by simple manipulation. Here the product is generalized to the real numbers, as can be found on pages 17 and 106 (all you need to read), the latter giving a formula for calculation. My question is whether a general solution to this equation can be found without the concepts of Summability Calculus because chances are you've never heard of it until seeing this post. Proving that this is the solution using this field of math is simple by turning to page 17 and using the differentiation rule for products. But I haven't taken a PDE course so I don't know any standard methods to solve this equation.

 

[anuttarasammyak]

The following PDE$$P^2\frac{\partial^2P}{\partial x\partial y}=(P^2+1)\frac{\partial P}{\partial x}\frac{\partial P}{\partial y}+P\frac{\partial^2P}{\partial y\partial x}$$can be solved with products,

As a layman, may I ask you what is the answer solved with products ?
Say
P(x,y)=X(x)Y(y)
The equation is
P(P^2-P+2)X'Y'=0
I don't think that I am on the right track.

 

[mathhabibi]

P(x,y)=X(x)Y(y)

Is not a double variable product because for there to be only two terms either $x=y+1$ or $y=x-1$. Good observation though.