Question

2z=(ax+y)^2+b

Answer 1

$$Solution$$

where a, b are arbitrary constants, is a complete integral of

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

$px+qy-q^2=\frac{1}{2}\frac{\delta((ax+y)^2+b)}{\delta x}x+\frac{1}{2}\frac{\delta((ax+y)^2+b)}{\delta y}y-(\frac{1}{2}\frac{\delta((ax+y)^2+b)}{\delta y})^2$

$\implies a(ax+y)x+(ax+y)y-(ax+y)^2=a^2x^2+axy+axy+y^2-(a^2x^2+2axy+y^2)=a^2x^2-a^2x^2+2axy-2axy+y^2-y^2=0+0+0=0$