Answer to Question #105030 in Differential Equations for Vaishali

Question

Find the surface which intersects the surfaces of the system z( x + y) = c(3z + 1) orthogonally and which passes through the circle x^2 + y^2 = 1 , z = 1

Answer 1

A one-parameter family of surfaces is characterized by the equation

$$f(x,y,z)={z(x+y) \over 3z+1}=c$$

We therefore have the linear partial differential equation

$${\partial f\over \partial x}\cdot{\partial z \over \partial x}+{\partial f \over \partial y}\cdot{\partial z\over \partial y}={\partial f\over \partial z}$$

$${z \over 3z+1}\cdot{\partial z \over \partial x}+{z \over 3z+1}\cdot{\partial z\over \partial y}={(x+y) \over (3z+1)^2}$$

$$z(3z+1)\cdot{\partial z \over \partial x}+z(3z+1)\cdot{\partial z\over \partial y}=x+y$$

The solution is

$$x-y=a, \ and\ \ x^2+y^2-2z^3-z^2=b$$

Thus any surface which is orthogonal to the given surfaces has equation of the form

$$x^2+y^2-2z^3-z^2=\psi(x-y)$$

For the particular surface passing through the circle $x^2+y^2=1, z=1$ we must take $\psi$ to be constant $-2$

$$\psi(x-y)=-2$$

The required surface is therefore

$$x^2+y^2=2z^3+z^2-2$$