Question

Find the differential equation of the family of surfaces φ[ z(x+y)² , x² ‐ y²] = 0

Answer 1

$\varphi (z(x+y)^2,x^2-y^2 )=0$

Therefore, $z(x+y)^2=C_1, \ \ x^2-y^2=C_2$

$z(x+y)^2=C_1 \ \Rightarrow 2z(x+y)dx +2z(x+y)dy+(x+y)^2dz=0,$

$2zdx+2zdy+(x+y)dz=0, \ \ 2z(dx+dy)=-(x+y)dz,\ \ \frac{dx+dy}{x+y}=-\frac{dz}{2z}$

$x^2-y^2=C_2 \ \Rightarrow 2xdx-2ydy=0, \ \ xdx=ydy, \ \ \frac{dx}{y}=\frac{dy}{x}$

$\frac{dx}{y}=\frac{dy}{x}=\frac{dx+dy}{y+x}=-\frac{dz }{2z}$

Answer: $\frac{dx}{y}=\frac{dy}{x}=\frac{dz }{-2z}$