Question

Show that 2z=(ax+y)^2+b,where a,b are arbitrary constants is a complete
integral of px+qy-q^2=0

Answer 1

$$2z=(ax+y)^2+b\rightarrow\boxed{z(x,y)=\frac{1}{2}(ax+y)^2+\frac{b}{2}}$$

Then,

$$p=\frac{\partial z}{\partial x}=\frac{1}{2}\cdot 2\cdot(ax+y)\cdot a=a^2x+ay\\[0.3cm] q=\frac{\partial z}{\partial y}=\frac{1}{2}\cdot 2\cdot(ax+y)=ax+y\\[0.3cm] px+qy-q^2=(a^2x+ay)x+(ax+y)y-(ax+y)^2=\\[0.3cm] =a^2x^2+axy+axy+y^2-(a^2x^2+2axy+y^2)=\\[0.3cm] =a^2x^2+2axy+y^2-a^2x^2-2axy-y^2=0$$

Conclusion,

$$\boxed{2z=(ax+y)^2+b\rightarrow px+qy-q^2=0}$$