To Solve: $\displaystyle \frac{\partial z}{\partial x} \left \{1-\left(\frac{\partial z}{\partial y}\right)^2\right \}=\frac{\partial z}{\partial y}(1-z)$
My Attempt:
Assume $z$ is a function of $\displaystyle u=x+ay$, where a is an arbitrary constant.
Therefore, $\displaystyle p= \frac{\partial z}{\partial x}=\frac{dz}{du}\frac{\partial u}{\partial x}=\frac{dz}{du}$
and $\displaystyle q= \frac{\partial z}{\partial y}=\frac{dz}{du}\frac{\partial u}{\partial y}=a\frac{dz}{du}$
$\displaystyle \frac{dz}{du}-a^2\left(\frac{dz}{du}\right)^3=a\frac{dz}{du}(1-z)$
When I cancel $\displaystyle \frac{dz}{du}$ from each side during simplication, is it justified that I am assuming that $\displaystyle \frac{dz}{du}\neq 0$?
I don't know how that assumption holds true. Please advise.
The PDE is of the form $f(z, p, q) =0$.
The first step is put $q=ap$ in the given equation.
The equation becomes $p(1-a^2p^2) =ap(1-z)$.
$1-a^2p^2=a-az$
$p=\frac{\sqrt{1-a+az}}{a}$
$dz=pdx+qdy$
$dz=pdx+apdy$
$dz=p(dx+ady)$
$\frac{dz}{p}=dx+ady$
$\frac{adz}{\sqrt{1-a+az}}=dx+ady$
Put $1-a+az=t$.
$adz=dt$
Integrating, we get
$2\sqrt{1-a+az}=x+ay+c$
Squaring both sides, the general solution is $4(1-a+az) =(x+ay+c)^2$.