The partial differential equation obtained by eliminating φ from :

φ(x + y + z, x² + y² − z²) = 0 is : 

Concept:

The given function is: \( \phi(u, v) = 0 \) where

\( u = x + y + z \) and \( v = x^2 + y^2 - z^2 \)

To eliminate \( \phi \), we use the method of characteristics by calculating the partial derivatives with respect to x, y, and z using the chain rule:

Calculation:

Let \( p = \frac{\partial z}{\partial x},\ q = \frac{\partial z}{\partial y} \)

Differentiate \( \phi(u,v) = 0 \) partially with respect to x:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial x}) + \frac{\partial \phi}{\partial v}(2x - 2z \cdot \frac{\partial z}{\partial x}) = 0 \)

Similarly, with respect to y:

\( \frac{\partial \phi}{\partial u}(1 + \frac{\partial z}{\partial y}) + \frac{\partial \phi}{\partial v}(2y - 2z \cdot \frac{\partial z}{\partial y}) = 0 \)

Let \( \phi_u = A,\ \phi_v = B \), then the two equations become:

\( A(1 + p) + B(2x - 2z p) = 0 \ \ \ \text{(1)} \)

\( A(1 + q) + B(2y - 2z q) = 0 \ \ \ \text{(2)} \)

Now eliminate A and B by cross-multiplying:

\( (1 + p)(2y - 2z q) = (1 + q)(2x - 2z p) \)

Simplify:

\( (1 + p)(2y - 2z q) - (1 + q)(2x - 2z p) = 0 \)

\( 2y + 2yp - 2zq - 2pzq - 2x - 2xq + 2zp + 2zpq = 0 \)

After simplification and rearranging terms, you arrive at:

\( (y + z)p + (x + z)q = x + y \)

Correct Option:

The correct answer is: 4) (y + z)p + (x + z)q = x + y