Find complete integral of $(y-x)(qy-px) = (p-q)^{2}$

Question

Find complete integral for partial differential equation $(y-x)(qy-px) = (p-q)^{2}$ where $p={ \partial z \over \partial x},q={ \partial z \over \partial y}$.

My attempt:

The given equation is f(x,y,z,p,q) = $(y-x)(qy-px) - (p-q)^{2}$ The Charpit's auxilary equation will be given by

$${dp \over 2px-(p+q)y}={dq \over 2qy-(p+q)x}={dx \over {2(p-q)-x2+xy}}={dy \over {-2(p-q)+xy-y2}}$$ From here I tried to proceed but no solvable fraction turned out.

Answer 1

Hint:

Let $\begin{cases}u=x+y\\v=x-y\end{cases}$ ,

Then $\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}=\dfrac{\partial z}{\partial u}+\dfrac{\partial z}{\partial v}$

$\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial y}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial y}=\dfrac{\partial z}{\partial u}-\dfrac{\partial z}{\partial v}$

$\therefore v\left(v\dfrac{\partial z}{\partial u}+u\dfrac{\partial z}{\partial v}\right)=4\left(\dfrac{\partial z}{\partial v}\right)^2$

Answer 2

Take $X$ and $Y$ to be new variables such that

$$X=x+y\tag1$$$$Y=xy\tag2$$

Given that $(y-x)(qy-px)=(p-q)^2$

Now, $$p=\dfrac{\partial z}{\partial x}=\dfrac{\partial z}{\partial X}\dfrac{\partial X}{\partial x}+\dfrac{\partial z}{\partial Y}\dfrac{\partial Y}{\partial x}=\dfrac{\partial z}{\partial X}+y\dfrac{\partial z}{\partial Y}\tag3$$and $$q=\dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial X}\dfrac{\partial X}{\partial y}+\dfrac{\partial z}{\partial Y}\dfrac{\partial Y}{\partial y}=\dfrac{\partial z}{\partial X}+x\dfrac{\partial z}{\partial Y}\tag4$$

Now plug in $p$ and $q$ in $(2)$ and we get $$(y-x)\left[y\left(\dfrac{\partial z}{\partial X}+x\dfrac{\partial z}{\partial Y}\right)-x\left(\dfrac{\partial z}{\partial X}+y\dfrac{\partial z}{\partial Y}\right)\right]=\left[\left(\dfrac{\partial z}{\partial X}+y\dfrac{\partial z}{\partial Y}\right)-\left(\dfrac{\partial z}{\partial X}+x\dfrac{\partial z}{\partial Y}\right)\right]^2$$or $$\implies(y-x)^2\dfrac{\partial z}{\partial X}=(y-x)^2\left(\dfrac{\partial z}{\partial Y}\right)$$ $$\implies \dfrac{\partial z}{\partial X}=\left(\dfrac{\partial z}{\partial Y}\right)^2\mbox{ or }\ \ \ p=q^2\tag5$$ where $p=\dfrac{\partial z}{\partial X}$ and $q=\dfrac{\partial z}{\partial Y}$

$(4)$ is of the form $f(p,q)=0$

So, $$z=aX+bY+c\tag6$$

where $a=b^2$, on putting $a$ for $p$ and $b$ for $q$ in $(5)$

Therefore, from $(6)$, the required complete integral is $$z=b^2X+bY+c$$or$$z=b^2(x+y)+bxy+c\ \ \mbox{ from }(6)$$