Charpit method: non-linear PDE I have a question: $$p^{2}x+q^{2}y = z.$$
I formed the Charpit auxiliary equation as follows
$$ \frac{\mathrm{d}x}{2px} = \frac{\mathrm{d}y}{2py} = \frac{\mathrm{d}z}{2(p^2x + q^2y)} = \frac{\mathrm{d}p}{p-p^2} = \frac{\mathrm{d}q}{q-q^2}.$$
After forming the equation I was unable to solve further (I applied everything I was taught). So I did some research (checked several books, and on the website as well) and found the next step to be
$$\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}. \tag{1}$$
After which I was able to solve. But I can not understand how they derived the relation (1). Can someone explain what method they applied here?
Disclaimer: The course we are being taught is engineering mathematics.

 A: Charpit's equation:
$$\boxed{\frac{dx}{2px} = \frac{dy}{\color{red}{2qy}} = \frac{dz}{2(p^2x + q^2y)} = \frac{dp}{p-p^2} = \frac{dq}{q-q^2}}$$
We have from Charpit's equation
$$\frac{dy}{2qy} =\frac{dq}{q-q^2}$$
Rearranging terms
$$(q-q^2){dy} =2qy{dq}$$
$$q{dy} =2qy{dq}+q^2dy$$
$$\text { (1) }\frac{dy}{qy} =\frac {2qy{dq}+q^2dy}{q^2y}$$
We have also from Charpit's equation that :
$$\frac{dx}{2px} = \frac{dp}{p-p^2}$$
$${dx}{(p-p^2)} = 2px{dp}$$
$${pdx} = 2px{dp}+p^2dx$$
$$\text { (2) }\frac {dx}{px} = \frac {2px{dp}+p^2dx} {p^2x}$$
We use a third equality from Charpit's equation
$$\frac {dx}{2px}=\frac{dy}{2qy} \implies \frac {dx}{px}=\frac{dy}{qy} $$
Therefore 
$$\boxed{\frac{p^2\,\mathrm{d}x + 2px\,\mathrm{d}p}{p^2x} = \frac{q^2\,\mathrm{d}y + 2qy\,\mathrm{d}q}{q^2y}}$$
That you can easily integrate...
A: Note that the middle term is equal to $\frac{dz}{2z}$ and use it as the reference value. Then
$$
\frac{p^2dx}{p^2x}=\frac{pdz}{z},\quad \frac{2px\,dp}{p^2x}=\frac{2dp}{p}=\frac{(1-p)\,dz}{z}
\implies \frac{d(p^2x)}{p^2x}=\frac{dz}{z}
$$
and similarly for $q^2y$.

Some other obvious simple steps are to recognize the decoupled nature of
$$
\frac{dz}{2z}=-\frac{d(1/p)}{1/p-1}=-\frac{d(1/q)}{1/q-1}
$$
which integrates to
$$
\frac12\ln|z|=c_1-\ln|1/p-1|=c_2-\ln|1/q-1|
\\\iff\\
\sqrt{|z|}=C_1\frac{p}{1-p}=C_2\frac{q}{1-q}
$$
Then insert
$$
p=\frac{\sqrt{|z|}}{C_1+\sqrt{|z|}},\qquad q=\frac{\sqrt{|z|}}{C_2+\sqrt{|z|}}
$$
into the first three relations
$$
\frac{dx}{2xp}=\frac{dy}{2yq}=\frac{dz}{2z}
$$
which should then be easy to integrate.

Or alternatively, combine the $x$ and $p$ fractions to 
$$
\frac{dx}{x}=\frac{2dp}{1-p}\implies x(1-p)^2=a,
$$
and similarly
$$
\frac{dy}{y}=\frac{2dq}{1-q}\implies y(1-q)^2=b.
$$
