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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.

square_one
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  • The factor $a$ is forgotten on the right-hand side in your third formula. And on the left-hand side in the second term as well. – Vladimir Jun 22 '14 at 09:00
  • The substitution $u=1-z$ will slightly simplify the PDE to $\dfrac{\partial u}{\partial x}\left(1-\left(\dfrac{\partial u}{\partial y}\right)^2\right)=u\dfrac{\partial u}{\partial y}$ . – doraemonpaul Jun 22 '14 at 19:32

1 Answers1

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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 $4(1-a+az) =(x+ay+c)^2$.

Gonçalo
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Anbu
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