Question

Eliminate the arbitrary constant a from 2z=(ax+y)^2

Answer 1

We will use the following standard notation to denote the partial derivatives:

$${\eth z \over \eth x}=p, {\eth z \over \eth y}=q$$

Given $2z=(ax+y)^2$

Differentiating the equation partially with respect to x and y respectively we get

$$2{\eth z \over \eth x}=2a(ax+y)$$

$$2{\eth z \over \eth y}=2(ax+y)$$

Then

$$p=a(ax+y), q=ax+y$$

We have

$$px+qy-q^2=ax(ax+y)+y(ax+y)-(ax+y)^2=$$

$$=(ax+y)^2-(ax+y)^2=0$$

We show that $2z=(ax+y)^2$, where $a$ is an arbitrary constant, is a complete integral of

$$px+qy-q^2=0$$

The partial differential equation is

$$2z=q^2$$