Answer to Question #233764 in Differential Equations for Shamim

Question

z=a(x+y)+b

Answer 1

This question is incomplete but we can treat this as a problem of elimination of the arbitrary constant.

$z=a(x+y)+b...(1) \\$

Differentiating (1) with respect to $x, y$ , we get

$p=\frac{\partial z}{\partial x}=\frac{\partial (a(x+y)+b)}{\partial x}=a...(2) \\p=\frac{\partial z}{\partial y}=\frac{\partial (a(x+y)+b)}{\partial y}=a...(3)$

Now, subtracting (3) from (2), we get

$p-q=0$